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11/16/11

So it seems that there are some math people around. To follow up on http://www.wallstreetoasis.com/forums/harder-brainteaser-than-the-coin-flip here is another one:

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

Show that:
a+b=c+d

Requires middle school math, but sure can ruin an interview.
As before, SBs for 1st correct answer.

Edit - all numbers are strictly positive. Apologies for confusion, but the problem remains open.

Edit - To provide an update, this has, thus far, been proven in 1 confirmed way by nonTargetChimp 9 (see his text file proof), and also in 1 more way that appears correct by unForseen (or rather by a friend of unForseen that works in HR). Another proof using trigonometry has been proposed and has not been fully evaluated but could be correct, and blastoise tried to prove this using fancy linear algebra but so far that proof does not appear to be correct, since he was able to prove it ignoring the positive constraint, meaning he proved something that is false. The most elegant solution to date, in my opinion, would have to be the HR rep's as presented by unForseen, which is found on page 8.

Background on the problem - came from an 8th grade city level math competition from the late 1960s at a mathematical school in the USSR.

Comments (399)

11/16/11

Just divide the 2nd equation by the first it's really simple.

(a^3 + b^3)/(a^2 + b^2) = (c^3 +d^3)/(c^2+b^2) <=> a+b=c+d

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In reply to sinthushan
11/16/11
sinthushan:

Just divide the 2nd equation by the first it's really simple.

(a^3 + b^3)/(a^2 + b^2) = (c^3 +d^3)/(c^2+b^2) <=> a+b=c+d

No such luck.
http://www.wolframalpha.com/input/?i=%28a^3+%2B+b^3%29%2F%28a^2+%2B+b^2%29

11/16/11

I just realized how stupid my answer was.

In reply to sinthushan
11/16/11
sinthushan:

Just divide the 2nd equation by the first it's really simple.

(a^3 + b^3)/(a^2 + b^2) = (c^3 +d^3)/(c^2+b^2) <=> a+b=c+d

Fail.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

11/16/11

why can't I just square root the 2nd equation? That's whats throwing me off right now. Are you sure the question doesn't ask you to find what a b c d are?

In reply to sinthushan
11/16/11
sinthushan:

why can't I just square root the 2nd equation? That's whats throwing me off right now. Are you sure the question doesn't ask you to find what a b c d are?

You can "square root" a perfect square, only. The question is correct as it is.

11/16/11

Take the square root of each variable?

I eat success for breakfast...with skim milk

In reply to TonyPerkis
11/16/11
TonyPerkis:

Take the square root of each variable?

doesn't work that way (a^2 + b^2)^1/2 doesn't equal a+b

11/16/11

Re-write 2nd equation as:

a^2 . a + b^2 . b = c^2 . c + d^2 . d

then divide it by 1st equation and end up with a+b = c+d ? lol

11/16/11

My guess (disclaimer: I am not mathematically inclined):

Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to Flake
11/16/11
Flake:

My guess (disclaimer: I am not mathematically inclined):

Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.

Your intuition may be correct but it is certainly not a proof.

In reply to Flake
11/16/11

Math Proofs....HUMBUG!!!

Get busy living

In reply to Dr Joe
11/16/11
Dr Joe:
Flake:

My guess (disclaimer: I am not mathematically inclined):

Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.

Your intuition may be correct but it is certainly not a proof.

I fail as well.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to Flake
11/16/11
Flake:
Dr Joe:
Flake:

My guess (disclaimer: I am not mathematically inclined):

Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.

Your intuition may be correct but it is certainly not a proof.

I fail as well.

SB for the chuckle

11/16/11

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

just guessing:

FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

divide both sides by (c^2 + d^2),

and end up with (a+b) = (c+d)

In reply to lookatmycock
11/16/11
lookatmycock:

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

Is my proof correct?

FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

divide both sides by (c^2 + d^2),

and end up with (a+b) = (c+d)

My GMAT level math tells me that's...kind of not right.

(a+b)(a^2 + b^2) = (c+d)(c^2 + d^2) does not equal a^3 + b^3 = c^3 + d^3.

I also don't get UFO's thing and how that proves a + b = c + d.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to lookatmycock
11/16/11
lookatmycock:

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

just guessing:

FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

divide both sides by (c^2 + d^2),

and end up with (a+b) = (c+d)

a^3 + b^3 does not equal (a+b)(a^2 + b^2)
(a+b)(a^2 + b^2) equals a^3 + 2ab^2 + b^3

11/16/11

The above two corrections are correct (and necessary). Also, all material here is within the scope of the GMAT, so if you prefer, think of it as a data sufficiency question:

Is a+b=c+d?
(1) a^2+b^2=c^2+d^2
(2) a^3+b^3=c^3+d^3

11/16/11

WHOOPS, edited...

a^2 + b^2 = c^2 + d^2

SO

a^2 + b^2 = a^2 + b^2

AND

a^3 + b^3 = a^3 + b^3

THEREFORE

a^x b^x = a^x b^x

SO

a^1 b^1 = a^1 + b^1

AND

a + b = a + b = c + d

a + b = c + d

Or something like that. Apparently, a 5th grader is smarter than me.

Get busy living

In reply to Flake
11/16/11
Flake:
lookatmycock:

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

Is my proof correct?

FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

divide both sides by (c^2 + d^2),

and end up with (a+b) = (c+d)

My GMAT level math tells me that's...kind of not right.

(a+b)(a^2 + b^2) = (c+d)(c^2 + d^2) does not equal a^3 + b^3 = c^3 + d^3.

I also don't get UFO's thing and how that proves a + b = c + d.

function yoFuckThisShit(www, forum, topic, x)
{

do while x > 0
{
yoFuckThisShit('WSO', 'monkeyingaround', 'Do You Know Algebra?', x)
msgBox.Display ('This shows college and grueling engineering psets didn't prep me for shit', alert);
x++
}

}

init_app()
{
this.yoFuckThisShit ('WSO', 'monkeyingaround', 'Do You Know Algebra?', 1);

}

11/16/11

Should I really embarrass everyone by doing this right now?

I'll tell you what, I solve this and someone gives me back the 1000 I just lost today on First Solar Calls

"Look, you're my best friend, so don't take this the wrong way. In twenty years, if you're still livin' here, comin' over to my house to watch the Patriots games, still workin' construction, I'll fuckin' kill you. That's not a threat, that's a fact.

11/16/11

C,

let a = 1 b = -1 c = 1 d = 1
a^2+b^2=c^2+d^2 holds true
a+b=c+d is not equal

let a = 1 b = 1 c = 1 d = 1

a^2+b^2=c^2+d^2 holds true
a+b=c+d is equal

1 is insufficient

same logic applies to 2

Only when combining can you lock down the sign so C

In reply to Dr Joe
11/16/11
Dr Joe:

The above two corrections are correct (and necessary). Also, all material here is within the scope of the GMAT, so if you prefer, think of it as a data sufficiency question:

Is a+b=c+d?
(1) a^2+b^2=c^2+d^2
(2) a^3+b^3=c^3+d^3

yeah my bad. i was thinking about (a+b)^3 lol

In reply to Will Hunting
11/16/11
Will Hunting:

Should I really embarrass everyone by doing this right now?

I'll tell you what, I solve this and someone gives me back the 1000 I just lost today on First Solar Calls

Please embarrass everyone by doing this right now.

In reply to FutureBanker09
11/16/11
FutureBanker09:

C,

let a = 1 b = -1 c = 1 d = 1
a^2+b^2=c^2+d^2 holds true
a+b=c+d is not equal

let a = 1 b = 1 c = 1 d = 1

a^2+b^2=c^2+d^2 holds true
a+b=c+d is equal

1 is insufficient

same logic applies to 2

Only when combining can you lock down the sign so C

Well you have proven its not A, B or D but you haven't really proven C.

In reply to FutureBanker09
11/16/11
FutureBanker09:

C,

let a = 1 b = -1 c = 1 d = 1
a^2+b^2=c^2+d^2 holds true
a+b=c+d is not equal

let a = 1 b = 1 c = 1 d = 1

a^2+b^2=c^2+d^2 holds true
a+b=c+d is equal

1 is insufficient

same logic applies to 2

Only when combining can you lock down the sign so C

That's what I was trying to say in my post with the whole plugging of the numbers thing and picking positive/negative signs.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to Dr Joe
11/16/11
Dr Joe:
FutureBanker09:

C,

let a = 1 b = -1 c = 1 d = 1
a^2+b^2=c^2+d^2 holds true
a+b=c+d is not equal

let a = 1 b = 1 c = 1 d = 1

a^2+b^2=c^2+d^2 holds true
a+b=c+d is equal

1 is insufficient

same logic applies to 2

Only when combining can you lock down the sign so C

Well you have proven its not A, B or D but you haven't really proven C.

Lol hoping you overlooked that

In reply to Will Hunting
11/16/11
Will Hunting:

Should I really embarrass everyone by doing this right now?

I'll tell you what, I solve this and someone gives me back the 1000 I just lost today on First Solar Calls

I give up do it

11/16/11

You guys realise that A^3+B^3 = (A+B )(A^2+B^2 - AB) right? not (A+B)(A^2+B^2) like i saw someone above doing

11/16/11

somebody go do it the long brute way, if this isn't a data sufficiency question:

first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

of course there's a shorter way but im an idiot today :D

In reply to Warhead
11/16/11
Warhead:

You guys realise that A^3+B^3 = (A+B )(A^2+B^2 - AB) right? not (A+B)(A^2+B^2) like i saw someone above doing

yeah i realized that too late lol. fail.

In reply to FutureBanker09
11/16/11
FutureBanker09:
lookatmycock:

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

just guessing:

FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

divide both sides by (c^2 + d^2),

and end up with (a+b) = (c+d)

a^3 + b^3 does not equal (a+b)(a^2 + b^2)
(a+b)(a^2 + b^2) equals a^3 + 2ab^2 + b^3

no it doesnt.

a+b(a^2 + b^2) = a^3 + ba^2 + ab^2 + b^ 3

In reply to Will Hunting
11/16/11
Will Hunting:

Should I really embarrass everyone by doing this right now?

I'll tell you what, I solve this and someone gives me back the 1000 I just lost today on First Solar Calls

haha do it. im curious. i like these math threads. shows how much of an idiot we can all be sometimes :D

In reply to lookatmycock
11/16/11
lookatmycock:

somebody go do it the long brute way, if this isn't a data sufficiency question:

first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

of course there's a shorter way but im an idiot today :D

plug all that crap in, solve, then plug in some more and see what you get

In reply to Dr Joe
11/16/11
Dr Joe:
lookatmycock:

somebody go do it the long brute way, if this isn't a data sufficiency question:

first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

of course there's a shorter way but im an idiot today :D

plug all that crap in, solve, then plug in some more and see what you get

shhhhh. im trying to make someone go through all that.

11/16/11

Let me give it a shot.

First, a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(c^2 - ab + d^2) [since a^2 + b^2 = c^2 + d^2].

Now, we know that a^3 + b^3 = c^3 + d^3, and so

(a+b)(c^2 - ab + d^2) = (c+d)(c^2 - cd + d^2)

For (a+b) = (c+d), we must show that ab = cd (this is a necessary condition).

If a+b = c+d, then

(a+b)^2 = a^2 + 2ab + b^2
(c+d)^2 = c^2 + 2cd + d^2

But we already know that a^2 + b^2 = c^2 + d^2, and so 2ab = 2cd. Hence if a+b = c+d (and our conditions hold) then ab = cd.

I don't know about you guys, but I have a life + work to do so if its wrong, so be it.

Just Do It

11/16/11
In reply to maximumlikelihood
11/16/11
maximumlikelihood:

Let me give it a shot.

First, a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(c^2 - ab + d^2) [since a^2 + b^2 = c^2 + d^2].

Now, we know that a^3 + b^3 = c^3 + d^3, and so

(a+b)(c^2 - ab + d^2) = (c+d)(c^2 - cd + d^2)

For (a+b) = (c+d), we must show that ab = cd (this is a necessary condition).

If a+b = c+d, then

(a+b)^2 = a^2 + 2ab + b^2
(c+d)^2 = c^2 + 2cd + d^2

But we already know that a^2 + b^2 = c^2 + d^2, and so 2ab = 2cd. Hence if a+b = c+d (and our conditions hold) then ab = cd.

I don't know about you guys, but I have a life + work to do so if its wrong, so be it.

Well you are the first one on the right track. You have shown that proving a+b=c+d is equivalent to proving ab=cd, which is correct. However, in proving 2ab=2cd, you implicitly assumed a+b=c+d, hence you proved that if a+b=c+d, then a+b=c+d, which is redundant. However, thanks for bringing a hint of actual math to this thread.

11/16/11

I'm pretty sure for the next part I have to cross multiply the two equations and show that ab = cd, but man what the hell I don't have time for this shit

Sorry about the redundant proof, I felt bad typing it in

Just Do It

11/16/11

Also, I just showed that if a+b = c+d, I will get this result because

(a+b)^2 = (c+d)^2, and a^2 + b^2 = c^2 + d^2.

Just Do It

In reply to maximumlikelihood
11/16/11
maximumlikelihood:

I'm pretty sure for the next part I have to cross multiply the two equations and show that ab = cd, but man what the hell I don't have time for this shit

Sorry about the redundant proof, I felt bad typing it in

The reason I posted this is that the actual proof is quite elegant and non-trivial, and cross-multiplying doesn't come close to cutting it. It's a good problem to solve in boring classes.

11/16/11

I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL

"Look, you're my best friend, so don't take this the wrong way. In twenty years, if you're still livin' here, comin' over to my house to watch the Patriots games, still workin' construction, I'll fuckin' kill you. That's not a threat, that's a fact.

In reply to Will Hunting
11/16/11
Will Hunting:

I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL

Factoring will not be sufficient here. You keep saying you can do it - we are all waiting with SBs in hand.

In reply to Dr Joe
11/16/11
Dr Joe:
Will Hunting:

I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL

Factoring will not be sufficient here. You keep saying you can do it - we are all waiting with SBs in hand.

I think he meant factorization. Anyway, I will solve it once I get out of class and have time to type.

"Look, you're my best friend, so don't take this the wrong way. In twenty years, if you're still livin' here, comin' over to my house to watch the Patriots games, still workin' construction, I'll fuckin' kill you. That's not a threat, that's a fact.

11/16/11

Is this doable by direct proof?
Or do we have to do RAA or something? When we get to the part where we have to show ab=cd

11/16/11

does it require using imaginary numbers, as in a^2 + b^2 = (a+bi)(a-bi)?

In reply to Warhead
11/16/11
Warhead:

Is this doable by direct proof?
Or do we have to do RAA or something? When we get to the part where we have to show ab=cd

You know, maybe, sort of. Something in the middle.

In reply to FutureBanker09
11/16/11
FutureBanker09:

does it require using imaginary numbers, as in a^2 + b^2 = (a+bi)(a-bi)?

No

11/16/11

Dr. Joe you're such a tease.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

11/16/11

This is probably stupid, but I go to a nontarget so give me abreak.

take the natural log of everything thus:

2lna + 2lnb = 2lnc + 2lnd

divde by 2

lna + lnb = lnc +lnd

raise e^ln. thus

a + b = c + d

In reply to pszonkadonk
11/16/11
pszonkadonk:

This is probably stupid, but I go to a nontarget so give me abreak.

take the natural log of everything thus:

2lna + 2lnb = 2lnc + 2lnd

divde by 2

lna + lnb = lnc +lnd

raise e^ln. thus

a + b = c + d

Look up log of sum formula. Also, you failed to use the cubic equation, hence any result can't be sufficient.

11/16/11

8---------------D (|) ratio in this thread is mean 0

11/16/11

a^2+b^2=c^2+d^2
+a^2+b^2=c^2+d^2
= a^2(1+a)+b^2(1+b)=c^2(1+c)+d^2(1+d)

and then somehow you eliminate a^2, b^2, c^2, d^2 through some bullshit because a^2+b^2=c^2+d^2, leaving (1+a)+(1+b)=(1+c)+(1+d), so a+b=c+d eliminating the ones...

Am I on the right track?

In reply to total
11/16/11
total:

a^2+b^2=c^2+d^2
+a^2+b^2=c^2+d^2
= a^2(1+a)+b^2(1+b)=c^2(1+c)+d^2(1+d)

and then somehow you eliminate a^2, b^2, c^2, d^2 through some bullshit because a^2+b^2=c^2+d^2, leaving (1+a)+(1+b)=(1+c)+(1+d), so a+b=c+d eliminating the ones...

Am I on the right track?

Not at all. Sorry.

11/16/11

Apparently the answer is no.

11/16/11
11/16/11

I got this on a math test once in high school. Me being the smart ass that I am wrote this.

Since we have 4 undefined variables I can assign any numeric value that I want to each of the variables. Therefore a,b,c,d = 1. Having set the numeric value of the variables. a+b=c+d. I acutally got full credit on that problem because the teacher had not designated that a,b,c,d are not all equal to one another.

Follow the shit your fellow monkeys say @shitWSOsays

Life is hard, it's even harder when you're stupid - John Wayne

11/16/11

I haven't solved the problem (it doesn't look that easy to me), but if this indeed is similar to most complex GMAT problems, you likely need to manipulate one of the formulas (likely the 2nd) in a way that allows you to use the first formula. For example, if you reduced a^3 + b^3 in a way that isolated (a^2+b^2), you could then replace a^2+b^2 with c^2+d^2 and work the formula from there. If that doesn't work, perhaps manipulate BOTH formulas...

I could be wrong, but some food for thought for those that are tackling this.

CompBanker

11/17/11

I've gone through this a few times now and each time I end up proving myself wrong.

Can we set a^2+b^2=c^2+d^2 = 1
and a^3+b^3=c^3+d^3 = 1

such that ac + bd = 0?

IFF this is true we need:

bc-ad = ?

So multiply (by c^2 + d^2) on both sides:

(a^2 + b^2)*(c^2 + d^2) = c^2 + d^2

(ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = 1

(ac + bd)^2 - 2acbd + (bc-ad)^2 +2abcd = 1

We know that ac + bd= 0

- 0 -2abcd + (bc-ad)^2 + 2abcd= 1

Now reduce similar terms:

- (bc-ad)^2 = 1

- bc - ad = 1

or:

- bc - ad = -1

accept this approach ignores the second equation.

(I love this stuff even when Im wrong)

Making money is art and working is art and good business is the best art - Andy Warhol

In reply to dwight schrute
11/17/11
dwight schrute:

I've gone through this a few times now and each time I end up proving myself wrong.

Can we set a^2+b^2=c^2+d^2 = 1
and a^3+b^3=c^3+d^3 = 1

such that ac + bd = 0?

IFF this is true we need:

bc-ad = ?

So multiply (by c^2 + d^2) on both sides:

(a^2 + b^2)*(c^2 + d^2) = c^2 + d^2

(ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = 1

(ac + bd)^2 - 2acbd + (bc-ad)^2 +2abcd = 1

We know that ac + bd= 0

- 0 -2abcd + (bc-ad)^2 + 2abcd= 1

Now reduce similar terms:

- (bc-ad)^2 = 1

- bc - ad = 1

or:

- bc - ad = -1

accept this approach ignores the second equation.

(I love this stuff even when Im wrong)

I am not sure I follow what you are doing above - and as you admit, without using the 2nd equation, a solution is not possible. Your 1st assumption that everything is equal to 1 is also overly constricting. .

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11/17/11

come on monkeys...don't make Fermat shift in his grave.LOL

11/17/11

a2+b2=c2+d2
a2-c2 = d2-b2
(a+c)(a-c) = (d+b)(d-b)

Do the same with a^3+b^3=c^3+d^3

then you get

a+c = d+b
and
a2 + ac + c2 = d2 + bd + b2

Sqaure both sides of
a+c = d+b
you get
ac = bd

substract a2 + ac + c2 = d2 + bd + b2 times 3
you get
a - c = d -b

a + b = c + d

(my math -> english is pretty horrible)

11/17/11

Say, in what domain are we solving this? I'll assume the reals, then the assumptions do not hold.
Take the way less general a=1, b=1.

c^2+d^2=2
c^3+d^3=2

So your assumption would imply that c+d=2, but one can find other c and d that solve the above system. Let u=c+d, so from the second:

(c+d)(c^2+d^2-cd)=2
u(2-cd)=2
cd=2-(2/u)

then adding twice to the first equation

c^2+2cd+d^2=2+4-(4/u)
u^2=6-(4/u)
u^3-6u+4=0

As expected, one solution is u=c+d=1+1=2. But the other roots

u(u-2)(u+2)-2u+4=0
(u-2)(u^2+2u-2)=0
are the roots of the quadratic:
u^2+2u-2=0

and those are -1+or-sqrt(3)

So, c+d can be equal to those numbers, and not just to 2=1+1=a+b.
For example:
c=(a+sqrt(a^2+4a))/2
d=(a+sqrt(a^2+4a))/2

where a=-1+sqrt(3)

one of the roots of the quad.

In reply to President
11/17/11
President:

a2+b2=c2+d2
a2-c2 = d2-b2
(a+c)(a-c) = (d+b)(d-b)

Do the same with a^3+b^3=c^3+d^3

then you get

a+c = d+b
and
a2 + ac + c2 = d2 + bd + b2

Sqaure both sides of
a+c = d+b
you get
ac = bd

substract a2 + ac + c2 = d2 + bd + b2 times 3
you get
a - c = d -b

a + b = c + d

(my math -> english is pretty horrible)

I don't follow what you did with the cubes to get a+c=d+b

In reply to wadtk
11/17/11
wadtk:

Say, in what domain are we solving this? I'll assume the reals, then the assumptions do not hold.
Take the way less general a=1, b=1.

c^2+d^2=2
c^3+d^3=2

So your assumption would imply that c+d=2, but one can find other c and d that solve the above system. Let u=c+d, so from the second:

(c+d)(c^2+d^2-cd)=2
u(2-cd)=2
cd=2-(2/u)

then adding twice to the first equation

c^2+2cd+d^2=2+4-(4/u)
u^2=6-(4/u)
u^3-6u+4=0

As expected, one solution is u=c+d=1+1=2. But the other roots

u(u-2)(u+2)-2u+4=0
(u-2)(u^2+2u-2)=0
are the roots of the quadratic:
u^2+2u-2=0

and those are -1+or-sqrt(3)

So, c+d can be equal to those numbers, and not just to 2=1+1=a+b.
For example:
c=(a+sqrt(a^2+4a))/2
d=(a+sqrt(a^2+4a))/2

where a=-1+sqrt(3)

one of the roots of the quad.

Sorry, don't have time right now to analyze line-by-line but the real domain is a fine assumption. But the last lines there suggest that c=d=/=1. However, I think its silly to argue that if c=d and a=b then the proposition doesn't hold -> you would have to imply that if 2a^2=2b^2 and 2a^3=2b^3 then a doesn't have to equal b, which is clearly absurd. I think you made a sign error somewhere, which, when resolved, will eliminate the contradiction, and the possibility of alternate solutions.

11/17/11

I (master of fin mathematics), and a team member (PhD. in Mathematics) stand by the above analysis. We want to be proved wrong, however....

11/17/11

Oh sh*t, it's going to be a showdown. MATH WARRRR!

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to wadtk
11/17/11
wadtk:

I (master of fin mathematics), and a team member (PhD. in Mathematics) stand by the above analysis. We want to be proved wrong, however....

Was I correct in inferring that c=d in your example? if so, how do you disprove my argument above for the absurdity of the results?

11/17/11

Too bad Will Hunting hasn't proved his self proclaimed math superiority.

11/17/11

President, you are assuming that xy = ab ==> x= a and y = b. This is not true.

11/17/11

U can proof it by simple induction, assuming positive integers. Look at your solutions of the 1st and 2nd equation. Reformulate them for i and i+1. Get rid of the i-th roots by taking all of it to the power of i. Then set a^i+b^i=c^i+d^i, plug in your solutions and u'll end up with 3a^i+3b^i=3c^i+3d^i and you're done. Also works for i+1...i+n. Should work at first sight, but no clue how to do it GMAT style ;)

11/17/11
11/17/11

a^2+b^2 = (a+b)^2
c^2+d^2= (c+d)^2
sqrt(a+b)^2= sqrt(c+d)^2
a+b=c+d

In reply to Dr Joe
11/17/11
Dr Joe:
wadtk:

I (master of fin mathematics), and a team member (PhD. in Mathematics) stand by the above analysis. We want to be proved wrong, however....

Was I correct in inferring that c=d in your example? if so, how do you disprove my argument above for the absurdity of the results?

what wadtk has shown is that if a=b=1 and assuming
c^2+d^2=2
and c^3 +d^3 = 2

then c+d = 2, 1+sqrt(3), 1-sqrt(3)

this proves than c+d is not necessarily equal to a+b and therefore proves that a+b = c+d is not always true even if c^2 + d^2 = a^2 + b^2
and c^3+d^3 = a^3 +b^3 are true

So the problem is not provable as it is not true!

In reply to JSC trainee
11/17/11
JSC trainee:

So the problem is not provable as it is not true!

I was wondering this same thing

Is there a math expert here who can give a definitive answer??

Get busy living

11/17/11

the problem does not hold for n^2; n>2 if abcd are all integers.

11/17/11

I believe wadtk has shown it to be unprovable, can anyone find a flaw in his/her reasoning?

11/17/11
bears1208:

Too bad Will Hunting hasn't proved his self proclaimed math superiority.

I actually lol at this and everyone was looking at me "wtf"?

@ Dr. Joe - is this some form of a diophantine equation? Maybe a hint is in order as apparently we are all too retarded to find a proof.

Making money is art and working is art and good business is the best art - Andy Warhol

In reply to JSC trainee
11/17/11
JSC trainee:

I believe wadtk has shown it to be unprovable, can anyone find a flaw in his/her reasoning?

I didn't do the excercise on paper. Just did some thinking about it, and if u look at my reasoning of above: if you follow it by induction, it should hold true for positive numbers. Looking at the generalized solutions of Eq. (1) or Eq. (2) you will get the roots with a negative sign, where wadtk is definitly right. Prooving it true for negative numbers shouldn't be possible.

11/17/11

This gave me a headache.

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In reply to sinthushan
11/17/11
sinthushan:

Just divide the 2nd equation by the first it's really simple.

(a^3 + b^3)/(a^2 + b^2) = (c^3 +d^3)/(c^2+b^2) <=> a+b=c+d

Your math is retarded. I guess you're ... a banker ?

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

In reply to shiggy
11/17/11
shiggy:

a^2+b^2 = (a+b)^2
c^2+d^2= (c+d)^2
sqrt(a+b)^2= sqrt(c+d)^2
a+b=c+d

Another high-school flunk out. Next..

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/17/11

My bad folks - all numbers are positive. For what it's worth, the example above was correct in that a possibility is A=1 B=1 C= -0.564579 D=1.296630. But all numbers are strictly positive.

11/17/11

It's ok, you got many people thinking about mathematics which is hall of fame in my book.

11/17/11

President is wrong, you don't get a+c = b+d, all you get is that
(a^2+ac+c^2)/(a+c ) = (b^2+bd+d^2)/(b+d)

11/17/11

okay so this isn't really that bad...

If I disagree with you, it's because you're wrong.

11/17/11

But the problem remains open...

11/17/11

a^2+b^2=c^2+d^2

think of c^2+d^2 as u

then a^2+b^2=u

then a, b, and radical u form a right triangle.

so use a triple like 3,4,5

a=3 b=4 and radical u=5

well if a=3 then a^2=9 and if b=4 then b^2=16 and radical u^2=25

radical u^2 is also equal to just "u" which is equal to c^2+d^2

so c^2+d^2=25

which makes sense because 9+16 = 25

no repeat the process with c,d, and 5 as a triangle.

you get c=3 and d=4

hence a+b=c+d

If I disagree with you, it's because you're wrong.

In reply to cold pizza 2
11/17/11

cold pizza, you have no knowledge of formal mathematical proofs, taking specific values is not a fucking proof.

In reply to cold pizza 2
11/17/11
cold pizza 2:

a^2+b^2=c^2+d^2

think of c^2+d^2 as u

then a^2+b^2=u

then a, b, and radical u form a right triangle.

so use a triple like 3,4,5

a=3 b=4 and radical u=5

well if a=3 then a^2=9 and if b=4 then b^2=16 and radical u^2=25

radical u^2 is also equal to just "u" which is equal to c^2+d^2

so c^2+d^2=25

which makes sense because 9+16 = 25

no repeat the process with c,d, and 5 as a triangle.

you get c=3 and d=4

hence a+b=c+d

Wtf is this shit, don't make any sense at all. Go back to high school pizza boy.

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/17/11

If I disagree with you, it's because you're wrong.

In reply to cold pizza 2
11/17/11
cold pizza 2:

here's a proof:
http://en.wikipedia.org/wiki/Pythagorean_theorem

Proof that you're retarded

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/17/11

lol

If I disagree with you, it's because you're wrong.

11/17/11

how about this:

a^2+b^2=c^2+d^2

say c^2+d^2 = z^2

so a^2+b^2 = z^2

then a, b, and z form a triangle where z is the hypotenuse

and c, d, and z form the same triangle. because if z=z and it is a right triangle according to pythagorean theorem, then the other sides are the same as well

still no good?

If I disagree with you, it's because you're wrong.

11/17/11

why do i need to use the second output if i can solve it with one?

If I disagree with you, it's because you're wrong.

In reply to cold pizza 2
11/17/11
cold pizza 2:

how about this:

a^2+b^2=c^2+d^2

say c^2+d^2 = z^2

so a^2+b^2 = z^2

then a, b, and z form a triangle where z is the hypotenuse

and c, d, and z form the same triangle. because if z=z and it is a right triangle according to pythagorean theorem, then the other sides are the same as well

still no good?

Hey buddy how about you try using the second output a^3+b^3=c^3+d^3
Nevermind it won't make any difference..

Troll ?

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

In reply to nonTargetChimp
11/17/11
nonTargetChimp:

Why don't you solve it for us all then, fucking genius?

I didn't say it was your fault, I said I was blaming you.

11/17/11

We know that: a^3+b^3 = c^3+d^3

Rewrite the equation so that it takes the form: (a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd) ---> call this equation 1

Now, if we can show that (a^2+b^2-ab)=(c^2+d^2-cd) then we will be able to divide them out of the equation, leaving us with our desired result, (a+b)=(c+d). So

Prove: (a^2+b^2-ab) = (c^2+d^2-cd) ----> call this equation 2

We know a^2+b^2=c^2+d^2, from the problem statement. We could subtract this value from both sides to show -ab = -cd and thus (a^2+b^2-ab)=(c^2+d^2-cd), but to show the step more clearly, let us say that:

k = (a^2+b^2) = (c^2+d^2)

plugging k into equation 2 yields: k-ab = k-cd

subtract k from both sides: -ab = -cd

Therefore, since (a^2+b^2) = (c^2+d^2) and -ab = -cd, we have proven that

(a^2+b^2-ab) = (c^2+d^2-cd). We could stop here, but to explicate it further, let j = (a^2+b^2-ab) = (c^2+d^2-cd).

Plug j into equation 1:

(a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd)
becomes
(a+b)(j) = (c+d)(j).

Divide by j.

(a+b) = (c+d).

QED, that's our proof.

I'll be here all week, feel free to leave tips or better yet, job offers.

11/17/11

^very nice

If I disagree with you, it's because you're wrong.

11/17/11

Nope. Equation 2 is what you need to prove but you use it to prove your conclusion.

In reply to Kolya
11/17/11
Kolya:

We know that: a^3+b^3 = c^3+d^3

Rewrite the equation so that it takes the form: (a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd) ---> call this equation 1

Now, if we can show that (a^2+b^2-ab)=(c^2+d^2-cd) then we will be able to divide them out of the equation, leaving us with our desired result, (a+b)=(c+d). So

Prove: (a^2+b^2-ab) = (c^2+d^2-cd) ----> call this equation 2

We know a^2+b^2=c^2+d^2, from the problem statement. We could subtract this value from both sides to show -ab = -cd and thus (a^2+b^2-ab)=(c^2+d^2-cd), but to show the step more clearly, let us say that:

k = (a^2+b^2) = (c^2+d^2)

plugging k into equation 2 yields: k-ab = k-cd

subtract k from both sides: -ab = -cd

Therefore, since (a^2+b^2) = (c^2+d^2) and -ab = -cd, we have proven that

(a^2+b^2-ab) = (c^2+d^2-cd). We could stop here, but to explicate it further, let j = (a^2+b^2-ab) = (c^2+d^2-cd).

Plug j into equation 1:

(a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd)
becomes
(a+b)(j) = (c+d)(j).

Divide by j.

(a+b) = (c+d).

QED, that's our proof.

I'll be here all week, feel free to leave tips or better yet, job offers.

I had the same thought process, but I haven't figured out how to prove ab=cd which is the core of this proof. You should check that part of your answer because it looks like an obvious circular argument to me.

11/17/11

Ouch, I just realized what I did. Give me a few minutes; I think I can get it.

11/17/11

Here's my attempt:

a^3+b^3=c^3+d^3
a^2+b^2=c^2+d^2

take the difference of the equations

(a^3-a^2)+(b^3-b^2)=(c^3-c^2)+(d^3-d^2)

for all positive integers, n^3-n^2=k is only true for n. So for that equation to be true, either a=c&b=d or a=d&b=c or a=b=c=d----->therefore a+b=c+d

11/18/11

Not limited to integers

11/18/11

Do we need to show "n^3-n^2=k is only true for n" more rigorously?

I could buy it, but I'm not sure without seeing a proof.

11/18/11

I guess n*n*n-n*n = k

turns to

n = k/n*n +1

so just one solution.

11/18/11

a^3+b^3 = c^3+d^3

a^3+b^3 = ab ((a^2)/b +(b^2)/a)

c^3+d^3 = cd ((c^2)/d +(d^2)/c)

ab ((a^2)/b +(b^2)/a) = cd ((c^2)/d +(d^2)/c)

-----------------------------------------------------------

a^2+b^2 = c^2+d^2

a^2+b^2 = ab (a/b +b/a)

c^2+d^2 = cd (c/d +d/c)

ab (a/b +b/a) = cd (c/d +d/c)

--------------------------------------

so, we got:

ab (a/b +b/a) = cd (c/d +d/c)

and

ab ((a^2)/b +(b^2)/a) = cd ((c^2)/d +(d^2)/c)

then

[ab (a/b +b/a) ] / (c/d +d/c) = cd

and

[ab ((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c) = cd

so

[ab (a/b +b/a) ] / (c/d +d/c) = [ab ((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c)

ab cancels out

(a/b +b/a) / (c/d +d/c) = [((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c)

let's say this entire thing = Z

so going back...

[ab ((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c) = cd = abZ

Do the same thing with:

ab (a/b +b/a) = cd (c/d +d/c)

and

ab ((a^2)/b +(b^2)/a) = cd ((c^2)/d +(d^2)/c)

to solve for ab

SO:

ab = [cd (c/d +d/c)] / (a/b +b/a)

ab = [cd ((c^2)/d +(d^2)/c)] / [((a^2)/b +(b^2)/a)]

cd (c/d +d/c)] / (a/b +b/a) = [cd ((c^2)/d +(d^2)/c)] / [((a^2)/b +(b^2)/a)]

cd cancels out

(c/d +d/c)] / (a/b +b/a) = [((c^2)/d +(d^2)/c)] / [((a^2)/b +(b^2)/a)]

let's say this entire thing equals Y

so going back...

ab = cdY

------------

now we have

ab=cdY and abZ=cd

cdY = cd/Z

cdYZ = cd

YZ=1

multiply both equations

ab * cd = cdY * abZ

SO ****************ab = cd ************

that shit is long though lol

If I disagree with you, it's because you're wrong.

11/18/11

ab = cd

going back to:

ab (a/b +b/a) = cd (c/d +d/c) - which came from the first equation

ab and cd cancel out

(a/b +b/a) =(c/d +d/c)

let's take out b and d

so b (a + 1/a) = d (c +1/c)

b (1) = d (1)

b = d

by that logic

a = c

so a+b = c+d

no?

If I disagree with you, it's because you're wrong.

11/18/11

^ wow that was retarded sorry forget that

If I disagree with you, it's because you're wrong.

In reply to Neighbor
11/18/11
tlynch5:
nonTargetChimp:

Why don't you solve it for us all then, fucking genius?

Ok, Sir.

Typing this is a pain so I will make (a2) = a^2

a2 + b2 = c2 + d2

=> (a2 + b2)^3 = (c2 + d2)^3
a6 + 3(a4)(b2) + 3(a2)(b4) + b6 = c6 + 3(c4)(d2) + 3(c2)(d4) + d6 (1)

a3 + b3 = c3 + d3

=> (a3 + b3)^2 = (c3 + d3)^2
a6 + 2(a3)(b3) + b6 = c6 + 2(c3)(d3) + d6 (2)

(1) - (2)
3(a4)(b2) + 3(a2)(b4) - 2(a3)(b3) = 3(c4)(d2) + 3(c2)(d4) - 2(c3)(d3)
a2b2(3a2 + 3b2 - 2ab) = c2d2(3c2 + 3d2 - 2cd)

Let ab = u, cd = v, a2 + b2 = c2 + d2 = k

So (u^2)(3k - 2u) = (v^2)(3k - 2v)
3k(u^2) - 2(u^3) = 3k(v^2) - 2(v^3)
3k(u^2 - v^2) - 2(u^3 - v^3) = 0
3k(u-v)(u+v) - 2(u-v)(u^2 + uv + v^2) = 0 (3)

Okay so:
Scenario 1: u - v = 0 => u = v => ab = cd. Done

Scenario 2: u - v =/= 0

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

3k = 3(a^2 + b^2) > 0
u^2 + uv + v^2 > 0
u + v = ab + cd > 0 (since a,b,c,d > 0)

=> Scenario 2 is impossible

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/18/11

dude

?

If I disagree with you, it's because you're wrong.

In reply to nonTargetChimp
11/18/11
nonTargetChimp:
tlynch5:
nonTargetChimp:

Why don't you solve it for us all then, fucking genius?

Ok, Sir.

Typing this is a pain so I will make (a2) = a^2

a2 + b2 = c2 + d2

=> (a2 + b2)^3 = (c2 + d2)^3
a6 + 3(a4)(b2) + 3(a2)(b4) + b6 = c6 + 3(c4)(d2) + 3(c2)(d4) + d6 (1)

a3 + b3 = c3 + d3

=> (a3 + b3)^2 = (c3 + d3)^2
a6 + 2(a3)(b3) + b6 = c6 + 2(c3)(d3) + d6 (2)

(1) - (2)
3(a4)(b2) + 3(a2)(b4) - 2(a3)(b3) = 3(c4)(d2) + 3(c2)(d4) - 2(c3)(d3)
a2b2(3a2 + 3b2 - 2ab) = c2d2(3c2 + 3d2 - 2cd)

Let ab = u, cd = v, a2 + b2 = c2 + d2 = k

So (u^2)(3k - 2u) = (v^2)(3k - 2v)
3k(u^2) - 2(u^3) = 3k(v^2) - 2(v^3)
3k(u^2 - v^2) - 2(u^3 - v^3) = 0
3k(u-v)(u+v) - 2(u-v)(u^2 + uv + v^2) = 0 (3)

Okay so:
Scenario 1: u - v = 0 => u = v => ab = cd. Done

Scenario 2: u - v =/= 0

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

3k = 3(a^2 + b^2) > 0
u^2 + uv + v^2 > 0
u + v = ab + cd > 0 (since a,b,c,d > 0)

=> Scenario 2 is impossible

nice!

In reply to nonTargetChimp
11/18/11
nonTargetChimp:
tlynch5:
nonTargetChimp:

Why don't you solve it for us all then, fucking genius?

Ok, Sir.

Typing this is a pain so I will make (a2) = a^2

a2 + b2 = c2 + d2

=> (a2 + b2)^3 = (c2 + d2)^3
a6 + 3(a4)(b2) + 3(a2)(b4) + b6 = c6 + 3(c4)(d2) + 3(c2)(d4) + d6 (1)

a3 + b3 = c3 + d3

=> (a3 + b3)^2 = (c3 + d3)^2
a6 + 2(a3)(b3) + b6 = c6 + 2(c3)(d3) + d6 (2)

(1) - (2)
3(a4)(b2) + 3(a2)(b4) - 2(a3)(b3) = 3(c4)(d2) + 3(c2)(d4) - 2(c3)(d3)
a2b2(3a2 + 3b2 - 2ab) = c2d2(3c2 + 3d2 - 2cd)

Let ab = u, cd = v, a2 + b2 = c2 + d2 = k

So (u^2)(3k - 2u) = (v^2)(3k - 2v)
3k(u^2) - 2(u^3) = 3k(v^2) - 2(v^3)
3k(u^2 - v^2) - 2(u^3 - v^3) = 0
3k(u-v)(u+v) - 2(u-v)(u^2 + uv + v^2) = 0 (3)

Okay so:
Scenario 1: u - v = 0 => u = v => ab = cd. Done

Scenario 2: u - v =/= 0

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

3k = 3(a^2 + b^2) > 0
u^2 + uv + v^2 > 0
u + v = ab + cd > 0 (since a,b,c,d > 0)

=> Scenario 2 is impossible

You forgot to change the sign in the following step:

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

In reply to cold pizza 2
11/18/11
cold pizza 2:

^ wow that was retarded sorry forget that

Do you mean forget both of your above posts, or do you still claim the really long one to work?

In reply to Dr Joe
11/18/11
Dr Joe:

You forgot to change the sign in the following step:

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

missed that...

still waiting on a proof!

In reply to jec
11/18/11
jec:
Kolya:

We know that: a^3+b^3 = c^3+d^3

Rewrite the equation so that it takes the form: (a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd) ---> call this equation 1

Now, if we can show that (a^2+b^2-ab)=(c^2+d^2-cd) then we will be able to divide them out of the equation, leaving us with our desired result, (a+b)=(c+d). So

Prove: (a^2+b^2-ab) = (c^2+d^2-cd) ----> call this equation 2

We know a^2+b^2=c^2+d^2, from the problem statement. We could subtract this value from both sides to show -ab = -cd and thus (a^2+b^2-ab)=(c^2+d^2-cd), but to show the step more clearly, let us say that:

k = (a^2+b^2) = (c^2+d^2)

plugging k into equation 2 yields: k-ab = k-cd

subtract k from both sides: -ab = -cd

Therefore, since (a^2+b^2) = (c^2+d^2) and -ab = -cd, we have proven that

(a^2+b^2-ab) = (c^2+d^2-cd). We could stop here, but to explicate it further, let j = (a^2+b^2-ab) = (c^2+d^2-cd).

Plug j into equation 1:

(a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd)
becomes
(a+b)(j) = (c+d)(j).

Divide by j.

(a+b) = (c+d).

QED, that's our proof.

I'll be here all week, feel free to leave tips or better yet, job offers.

I had the same thought process, but I haven't figured out how to prove ab=cd which is the core of this proof. You should check that part of your answer because it looks like an obvious circular argument to me.

yup this just needs a proof that ab=cd and it would work.

Money Never Sleeps? More like Money Never SUCKS amirite?!?!?!?

11/18/11

Wow still not solved? I'll really give it ago in a bit. I'm starting to lose my respect for WSO after some of the brainteaser answers I've seen...

Like that guy who didn't know the sum of 1 to 100...

Just Do It

In reply to maximumlikelihood
11/18/11
maximumlikelihood:

Wow still not solved? I'll really give it ago in a bit. I'm starting to lose my respect for WSO after some of the brainteaser answers I've seen...

Like that guy who didn't know the sum of 1 to 100...

Not knowing how to add 1 to 100, is inexcusable unless you were dropped on your head as a kid. However this problem is legitimately tough, I got a 50 on the Math on the GMAT, but have no clue how to solve this.

In reply to FutureBanker09
11/18/11
FutureBanker09:
maximumlikelihood:

Wow still not solved? I'll really give it ago in a bit. I'm starting to lose my respect for WSO after some of the brainteaser answers I've seen...

Like that guy who didn't know the sum of 1 to 100...

Not knowing how to add 1 to 100, is inexcusable unless you were dropped on your head as a kid. However this problem is legitimately tough, I got a 50 on the Math on the GMAT, but have no clue how to solve this.

^^^

MAXIMUM...how about YOU solve this???

Get busy living

In reply to Dr Joe
11/18/11
Dr Joe:
cold pizza 2:

^ wow that was retarded sorry forget that

Do you mean forget both of your above posts, or do you still claim the really long one to work?

i still claim the really long one (ab=cd)

just don't know how to use that to prove a+b=c+d

although i can prove a^4+b^4=c^4+d^4

so i'm retarded

but I also stand by the triangle proof. if this was a data sufficiency problem, you only need the first equation.

If I disagree with you, it's because you're wrong.

11/18/11

U cannot do the triangle proof, as u cannot solve it for i>2. If you can do it for i = 4 u should normally be able to generalize it for i and i +1 and u will be fine - as long as it holds.

In reply to cold pizza 2
11/18/11
cold pizza 2:
Dr Joe:
cold pizza 2:

^ wow that was retarded sorry forget that

Do you mean forget both of your above posts, or do you still claim the really long one to work?

i still claim the really long one (ab=cd)

just don't know how to use that to prove a+b=c+d

although i can prove a^4+b^4=c^4+d^4

so i'm retarded

but I also stand by the triangle proof. if this was a data sufficiency problem, you only need the first equation.

Don't have time to sort out the long proof now, but if you can show ab=cd, then proving a+b=c+d is trivial by expanding (a+b)^2 and using a^2+b^2=c^2+d^2. Also, without using the cubic equation, providing a counter-example is easy: a=1 b=2 c=sqrt(2) d=sqrt(3) for example so it can't hold so triangle goes out the window.

So if this was a data sufficiency question, the 1st is certainly not sufficient.

In reply to is-t
11/18/11
is-t:
Dr Joe:

You forgot to change the sign in the following step:

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

missed that...

still waiting on a proof!

You're right. Let me fix that

3k(u+v) - 2(u^2 + uv + v^2) = 0

Let u + v = s, uv = p

3k.s - 2.(s^2 - p) = 0
-2.s^2 + 3k.s + 2p = 0 (*)
Consider s the variable, recall that s = u + v = ab + cd > 0. So we need a positive solution for (*)

delta = 9.k^2 + 16.p > 0 (4)

Product of the two solutions = - 2.2p = -4. p < 0 (5)

So we definitely have at least one positive solution for (*), because (4) shows that we have 2 solutions for (*) and (5) shows that the two solutions have reverse signs. So scenario (2) is definitely possible.
Which leads to the conclusion that...
- Dr. Joe, your inputs are wrong again
- Troll !?!?!?!?!?!?!?!

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/18/11

a^2+b^2=a^3d^2+b^3+c^2 = (c+d)(c^2-cd+d^2)

Based on Fermats Theory
(ad)^e + (bd)^e = (cd)^e

In reply to nonTargetChimp
11/18/11
nonTargetChimp:
is-t:
Dr Joe:

You forgot to change the sign in the following step:

Divide both sides of (3) by (u-v)
3k(u+v) -2(u^2 + uv + v^2) = 0
3k = -2(u^2 + uv + v^2)/(u+v)

missed that...

still waiting on a proof!

You're right. Let me fix that

3k(u+v) - 2(u^2 + uv + v^2) = 0

Let u + v = s, uv = p

3k.s - 2.(s^2 - p) = 0
-2.s^2 + 3k.s + 2p = 0 (*)
Consider s the variable, recall that s = u + v = ab + cd > 0. So we need a positive solution for (*)

delta = 9.k^2 + 16.p > 0 (4)

Product of the two solutions = - 2.2p = -4. p < 0 (5)

So we definitely have at least one positive solution for (*), because (4) shows that we have 2 solutions for (*) and (5) shows that the two solutions have reverse signs. So scenario (2) is definitely possible.
Which leads to the conclusion that...
- Dr. Joe, your inputs are wrong again
- Troll !?!?!?!?!?!?!?!

Haha if true, this would be some good trolling lol.

But its not. Remember, if you think it doesn't hold, proving that you are right is very easy. Simply provide a set of values that prove me wrong. Proving that it does hold is (evidently) much tougher.

I am having trouble digging into your proof because I gave you a b c d and your proof about a b c d is expressed in terms of u, v, t, s, p, and k.

Perhaps its time for a hint.

In reply to Dr Joe
11/18/11
Dr Joe:
cold pizza 2:
Dr Joe:
cold pizza 2:

^ wow that was retarded sorry forget that

Do you mean forget both of your above posts, or do you still claim the really long one to work?

i still claim the really long one (ab=cd)

just don't know how to use that to prove a+b=c+d

although i can prove a^4+b^4=c^4+d^4

so i'm retarded

but I also stand by the triangle proof. if this was a data sufficiency problem, you only need the first equation.

Don't have time to sort out the long proof now, but if you can show ab=cd, then proving a+b=c+d is trivial by expanding (a+b)^2 and using a^2+b^2=c^2+d^2. Also, without using the cubic equation, providing a counter-example is easy: a=1 b=2 c=sqrt(2) d=sqrt(3) for example so it can't hold so triangle goes out the window.

So if this was a data sufficiency question, the 1st is certainly not sufficient.

don't even sort it out - it's wrong too. lol don't know what i was thinking

If I disagree with you, it's because you're wrong.

In reply to Dr Joe
11/18/11

Haha if true, this would be some good trolling lol.

This kind of trolling deserves capital punishment in a Nerdsville bro :)

Finding a numerical solution that can disprove your thesis might be out of my reach. You did not really prove me wrong though.

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

In reply to nonTargetChimp
11/18/11
nonTargetChimp:

Haha if true, this would be some good trolling lol.

This kind of trolling deserves capital punishment in a Nerdsville bro :)

Finding a numerical solution that can disprove your thesis might be out of my reach. You did not really prove me wrong though.

Yeah I know I didn't prove you wrong. But the thing is that your "conclusion" is that I am wrong. To prove your conclusion wrong, I would have to prove myself right. Which would mean posting a solution.

That being said, I am not sure your logic is all that sound in this instance. I understand your argument as "If Z=/=0, then I can divide through by it. However, if this later results in a contradiction, then Z=0 because the contradiction came from somewhere." What you have done thus far is that you have not found a contradiction, and concluded that since no contradiction, then Z=/=0. However, as you know, absence of evidence is not evidence of absence, and proving the system consistent would require going back to a b c d, which would be equivalent of finding a numerical solution.

11/18/11

He's like Yoda or something.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to Dr Joe
11/18/11
Dr Joe:
nonTargetChimp:

Haha if true, this would be some good trolling lol.

This kind of trolling deserves capital punishment in a Nerdsville bro :)

Finding a numerical solution that can disprove your thesis might be out of my reach. You did not really prove me wrong though.

Yeah I know I didn't prove you wrong. But the thing is that your "conclusion" is that I am wrong. To prove your conclusion wrong, I would have to prove myself right. Which would mean posting a solution.

That being said, I am not sure your logic is all that sound in this instance. I understand your argument as "If Z=/=0, then I can divide through by it. However, if this later results in a contradiction, then Z=0 because the contradiction came from somewhere." What you have done thus far is that you have not found a contradiction, and concluded that since no contradiction, then Z=/=0. However, as you know, absence of evidence is not evidence of absence, and proving the system consistent would require going back to a b c d, which would be equivalent of finding a numerical solution.

Okay, you're right, my proof might not be definitive yet, even I myself might be lost among the variables that I created. I could run an Excel model (yup that is the end of my technical competency) to (maybe) find a numerical solution but I don't have time right now.

Btw is your proof short and elegant ?

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

In reply to nonTargetChimp
11/18/11
nonTargetChimp:
Dr Joe:
nonTargetChimp:

Haha if true, this would be some good trolling lol.

This kind of trolling deserves capital punishment in a Nerdsville bro :)

Finding a numerical solution that can disprove your thesis might be out of my reach. You did not really prove me wrong though.

Yeah I know I didn't prove you wrong. But the thing is that your "conclusion" is that I am wrong. To prove your conclusion wrong, I would have to prove myself right. Which would mean posting a solution.

That being said, I am not sure your logic is all that sound in this instance. I understand your argument as "If Z=/=0, then I can divide through by it. However, if this later results in a contradiction, then Z=0 because the contradiction came from somewhere." What you have done thus far is that you have not found a contradiction, and concluded that since no contradiction, then Z=/=0. However, as you know, absence of evidence is not evidence of absence, and proving the system consistent would require going back to a b c d, which would be equivalent of finding a numerical solution.

Okay, you're right, my proof might not be definitive yet, even I myself might be lost among the variables that I created. I could run an Excel model (yup that is the end of my technical competency) to (maybe) find a numerical solution but I don't have time right now.

Btw is your proof short and elegant ?

Providing description of the proof may be equivalent to giving a hint. PM me if interested - don't want to post hints here. Unless you guys want them...

11/18/11

No hints lol I wanna see how long this goes on

I didn't say it was your fault, I said I was blaming you.

11/18/11

I got something short and elegant for you.

Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

In reply to Flake
11/18/11
Flake:

I got something short and elegant for you.

No homo..

Okay I throw in the towel. A dude with a Math Fin degree and his bud who got a PhD Math came here before so who am I to speak here, just a random non-target chimp..

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/18/11

B = A, C = A, D=A+B-C A=1 B=1 C=1 D=1+1-1=1

D^2 = A^2 + B^2 - C^2 =1^2 + 1^2 - 1^2

D^3 = A^3 + B^3-C^3 = 1^3 = 1^3+1^3-1^3=1^3

B= -A + C +D = 1= -1+1+1

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
1^2+1^2=1^2+1^2= (1+1)(1^2-1+1^2)=(1+1)(1^2-1+1^2)=2=2

That actually was a lot easier than I originally thought.

In reply to Im not very creative
11/18/11
Im not very creative:

B = A, C = A, D=A+B-C A=1 B=1 C=1 D=1+1-1=1

D^2 = A^2 + B^2 - C^2 =1^2 + 1^2 - 1^2

D^3 = A^3 + B^3-C^3 = 1^3 = 1^3+1^3-1^3=1^3

B= -A + C +D = 1= -1+1+1

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
1^2+1^2=1^2+1^2= (1+1)(1^2-1+1^2)=(1+1)(1^2-1+1^2)=2=2

That actually was a lot easier than I originally thought.

We have a winner, folks. Congrats, very elegant!

11/18/11

Someone else proved ab=cd so I'll start there. Could be wrong as I haven't taken a math class in almost a decade.

(1) Given: ab = cd
So a = cd/b and c = ab/d

(2) a^2 + b^2 = c^2 + d^2
Sub in for a and c from (1)
(c^2 * d^2)/b^2 + b^2 = (a^2 * b^2)/d^2 + d^2

(3) multiply each side b^2 * d^2 then rearrange and factor
(c^2 * d^4) + (b^4 * d^2) = (a^2 * b^4) + (b^2 * d^4)
(c^2 * d^4) - (b^2 * d^4) = (a^2 * b^4) - (b^4 * d^2)
d^4 * (c^2 - b^2) = b^4 * (a^2 - d^2)

(4) Rearrange a^2 + b^2 = c^2 + d^2
So c^2 - b^2 = a^2 - d^2 = X

(5) Sub in X in equation (3)
d^4 * X = b^4 *X
b^4 = d^4 so b=d for all positive numbers

(6) ab=cd and b=d, then a =c so a+b=c+d

That killed some time at work, almost time to start drinking.

In reply to Dr Joe
11/18/11
Dr Joe:
Im not very creative:

.....

We have a winner, folks. Congrats, very elegant!

That was actually kinda the direction I showed you one page before. It's pretty straight-forward if you do it in a generalized way. This case is i =1 (like Im not very creative showed). If you replace the 1's with i and you can show it another time for i+1 (what you can do) you have a mathematically nice and sound proof (of course not GMAT plug-in style). The trick is really just to calculate the positive solution for D, C, B, and A and then move forward (actually with i, i+1 it's the same and also pretty easy). Dunno, we had to do these induction proofs all the time during high school. I guess there's more exciting stuff at this age.

In reply to Dr Joe
11/18/11
Dr Joe:
Im not very creative:

B = A, C = A, D=A+B-C A=1 B=1 C=1 D=1+1-1=1

D^2 = A^2 + B^2 - C^2 =1^2 + 1^2 - 1^2

D^3 = A^3 + B^3-C^3 = 1^3 = 1^3+1^3-1^3=1^3

B= -A + C +D = 1= -1+1+1

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
1^2+1^2=1^2+1^2= (1+1)(1^2-1+1^2)=(1+1)(1^2-1+1^2)=2=2

That actually was a lot easier than I originally thought.

We have a winner, folks. Congrats, very elegant!

I don't follow how this is proof. All it shows is that it works for a=b=c?

In reply to Dr Joe
11/18/11
Dr Joe:
Im not very creative:

B = A, C = A, D=A+B-C A=1 B=1 C=1 D=1+1-1=1

D^2 = A^2 + B^2 - C^2 =1^2 + 1^2 - 1^2

D^3 = A^3 + B^3-C^3 = 1^3 = 1^3+1^3-1^3=1^3

B= -A + C +D = 1= -1+1+1

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
1^2+1^2=1^2+1^2= (1+1)(1^2-1+1^2)=(1+1)(1^2-1+1^2)=2=2

That actually was a lot easier than I originally thought.

We have a winner, folks. Congrats, very elegant!

You're fucking with us right Dr. Joe?

Just Do It

In reply to maximumlikelihood
11/18/11
maximumlikelihood:
Dr Joe:
Im not very creative:

B = A, C = A, D=A+B-C A=1 B=1 C=1 D=1+1-1=1

D^2 = A^2 + B^2 - C^2 =1^2 + 1^2 - 1^2

D^3 = A^3 + B^3-C^3 = 1^3 = 1^3+1^3-1^3=1^3

B= -A + C +D = 1= -1+1+1

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
1^2+1^2=1^2+1^2= (1+1)(1^2-1+1^2)=(1+1)(1^2-1+1^2)=2=2

That actually was a lot easier than I originally thought.

We have a winner, folks. Congrats, very elegant!

You're fucking with us right Dr. Joe?

Given A = B = C = D = n
And n = 1
Is true

What about n =/= 1?

Get busy living

In reply to occupy this
11/18/11
occupy this:

Someone else proved ab=cd so I'll start there. Could be wrong as I haven't taken a math class in almost a decade.

(1) Given: ab = cd
So a = cd/b and c = ab/d

(2) a^2 + b^2 = c^2 + d^2
Sub in for a and c from (1)
(c^2 * d^2)/b^2 + b^2 = (a^2 * b^2)/d^2 + d^2

(3) multiply each side b^2 * d^2 then rearrange and factor
(c^2 * d^4) + (b^4 * d^2) = (a^2 * b^4) + (b^2 * d^4)
(c^2 * d^4) - (b^2 * d^4) = (a^2 * b^4) - (b^4 * d^2)
d^4 * (c^2 - b^2) = b^4 * (a^2 - d^2)

(4) Rearrange a^2 + b^2 = c^2 + d^2
So c^2 - b^2 = a^2 - d^2 = X

(5) Sub in X in equation (3)
d^4 * X = b^4 *X
b^4 = d^4 so b=d for all positive numbers

(6) ab=cd and b=d, then a =c so a+b=c+d

That killed some time at work, almost time to start drinking.

You have it backwards - it has been repeatedly proven that if ab=cd then we are all set. The unsolved part is proving ab=cd.

Side note - looks like I got someone to join WSO just to solve math problems.

11/18/11

Lol sarcasm doesn't seem to come across very well online. I just picked the most nonsensical thing posted thus far and declared it true.

Edit - I can't believe I just had to clarify that

11/18/11

nonTargetChimp, I came up with the exact same solution as you, but it is wrong because a,b,c, and d don't have to be positive.

11/18/11

B=A C=A D=A+B-C A=5 B=5 C=5 D=5+5-5=5

D^2 = A^2 +B^2 - C^2 = 25=25+25-25

D^3 = A^3 +B^3 -C^3 = 125=125+125-125

B= -A + C + D = -5+5+5=5

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
= (5+5)(5^2-25+5^2)=250=(5+5)(5^2-25+5^2)=250 Which means that 5^3+5^3=5^3+5^3 or 125+125=125+125

For n =/=1

In reply to Dr Joe
11/18/11
Dr Joe:

Lol sarcasm doesn't seem to come across very well online. I just picked the most nonsensical thing posted thus far and declared it true.

Edit - I can't believe I just had to clarify that

LOL

You definitely had me until this post. I was like, wtf!?!?!?

If I disagree with you, it's because you're wrong.

In reply to Im not very creative
11/18/11
Im not very creative:

B=A C=A D=A+B-C A=5 B=5 C=5 D=5+5-5=5

D^2 = A^2 +B^2 - C^2 = 25=25+25-25

D^3 = A^3 +B^3 -C^3 = 125=125+125-125

B= -A + C + D = -5+5+5=5

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
= (5+5)(5^2-25+5^2)=250=(5+5)(5^2-25+5^2)=250 Which means that 5^3+5^3=5^3+5^3 or 125+125=125+125

For n =/=1

I am sorry but I really don't know what you are trying to say here. Let's start at the very beginning: why is C=A? What are you trying to prove?

In reply to the_red_baron
11/18/11
the_red_baron:

nonTargetChimp, I came up with the exact same solution as you, but it is wrong because a,b,c, and d don't have to be positive.

Yes they do - they are all positive.

11/18/11

Ok Dr. Joe,

I'm going to publicly ask for a hint. Lets be honest, getting to ab=cd is piss easy, its the next part that needs some direction.

I just don't want to spend hours on it on a Friday then find out some smartass has posted the proof. i.e. I'm getting paranoid and continuously refreshing this page.

Edit: Can you PM me the hint? Since now everyone knows. I'll obviously mention it in my proof if I get there.

Just Do It

11/18/11

I'm surprised nobody posted it on a math forum and got it solved in a matter of minutes...

In reply to magnum
11/18/11
magnum:

I'm surprised nobody posted it on a math forum and got it solved in a matter of minutes...

I was actually thinking the same thing. Though I am not sure it would be solved in a matter of minutes. I think we've had some fairly mathematically inclined individuals here as well.

In reply to Dr Joe
11/18/11
Dr Joe:
Im not very creative:

B=A C=A D=A+B-C A=5 B=5 C=5 D=5+5-5=5

D^2 = A^2 +B^2 - C^2 = 25=25+25-25

D^3 = A^3 +B^3 -C^3 = 125=125+125-125

B= -A + C + D = -5+5+5=5

A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)=
= (5+5)(5^2-25+5^2)=250=(5+5)(5^2-25+5^2)=250 Which means that 5^3+5^3=5^3+5^3 or 125+125=125+125

For n =/=1

I am sorry but I really don't know what you are trying to say here. Let's start at the very beginning: why is C=A? What are you trying to prove?

C does not always have to equal A...but it will be equal to either A or B...

11/18/11

Let a^2+b^2=c^2+d^2 = k

a + b = u, c + d = v

a^3+b^3=c^3+d^3

(a + b)(a^2+b^2 - ab) = (c + d)(c^2+d^2 - cd) (1)

Notice ab = [(a+b)^2 - a^2 - b^2]/2 = (u^2 - k)/2
Same with cd = (v^2 - k)/2

(1) u(k - (u^2 - k)/2) = v(k - (v^2 - k)/2)

Multiply both sides by 2
u.(2k - u^2 + k) = v(2k - v^2 + k)
u.(3k - u^2) = v.(3k - v^2)
3k(u-v) - (u^3 - v^3) = 0

Case 1: u = v. Done

Case 2: 3k = u^2 + v^2 - uv
or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)

If a + b > c + d
(2)
=> (a - b)^2 + (a+b)(c+d) - 2cd = 0

a + b > c + d
(a+b)(c+d) > (c + d)^2

=> Left side > (a - b)^2 + (c + d)^2 - 2cd = (a - b)^2 + c^2+d^2 > 0

If a + b

(2)
=> (c - d)^2 + (a+b)(c+d) - 2ab = 0

c + d > a + b
(a+b)(c+d) > (a + b)^2

=> Left side > (c - d)^2 + (a + b)^2 - 2ab = (c - d)^2 + a^2+b^2 > 0

=> Case 2 is impossible

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

In reply to nonTargetChimp
11/18/11
nonTargetChimp:

Let a^2+b^2=c^2+d^2 = k

a + b = u, c + d = v

a^3+b^3=c^3+d^3

(a + b)(a^2+b^2 - ab) = (c + d)(c^2+d^2 - cd) (1)

Notice ab = [(a+b)^2 - a^2 - b^2]/2 = (u^2 - k)/2
Same with cd = (v^2 - k)/2

(1) <=> u(k - (u^2 - k)/2) = v(k - (v^2 - k)/2)

Multiply both sides by 2
u.(2k - u^2 + k) = v(2k - v^2 + k)
u.(3k - u^2) = v.(3k - v^2)
3k(u-v) - (u^3 - v^3) = 0

Case 1: u = v. Done

Case 2: 3k = u^2 + v^2 - uv
or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)

If a + b > c + d
(2)
=> (a - b)^2 + (a+b)(c+d) - 2cd = 0

a + b > c + d
(a+b)(c+d) > (c + d)^2

=> Left side > (a - b)^2 + (c + d)^2 - 2cd = (a - b)^2 + c^2+d^2 > 0

If a + b < c + d

(2)
=> (c - d)^2 + (a+b)(c+d) - 2ab = 0

c + d > a + b
(a+b)(c+d) > (a + b)^2

=> Left side > (c - d)^2 + (a + b)^2 - 2ab = (c - d)^2 + a^2+b^2 > 0

=> Case 2 is impossible

edit:delete (my post)

In reply to nonTargetChimp
11/18/11
nonTargetChimp:

Let a^2+b^2=c^2+d^2 = k

a + b = u, c + d = v

a^3+b^3=c^3+d^3

(a + b)(a^2+b^2 - ab) = (c + d)(c^2+d^2 - cd) (1)

Notice ab = [(a+b)^2 - a^2 - b^2]/2 = (u^2 - k)/2
Same with cd = (v^2 - k)/2

(1) <=> u(k - (u^2 - k)/2) = v(k - (v^2 - k)/2)

Multiply both sides by 2
u.(2k - u^2 + k) = v(2k - v^2 + k)
u.(3k - u^2) = v.(3k - v^2)
3k(u-v) - (u^3 - v^3) = 0

Case 1: u = v. Done

Case 2: 3k = u^2 + v^2 - uv
or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)

If a + b > c + d
(2)
=> (a - b)^2 + (a+b)(c+d) - 2cd = 0

a + b > c + d
(a+b)(c+d) > (c + d)^2

=> Left side > (a - b)^2 + (c + d)^2 - 2cd = (a - b)^2 + c^2+d^2 > 0

If a + b < c + d

(2)
=> (c - d)^2 + (a+b)(c+d) - 2ab = 0

c + d > a + b
(a+b)(c+d) > (a + b)^2

=> Left side > (c - d)^2 + (a + b)^2 - 2ab = (c - d)^2 + a^2+b^2 > 0

=> Case 2 is impossible

Equation 2 seems wrong

Case 2: 3k = u^2 + v^2 - uv
or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)

Where did the 3 go? Coefficient of the squares should be 2, no?

11/18/11

i didn't come from a private school or anything, so maybe that's why, but...

none of this shit seems to me like middle school or even high school level math.

what is this "i = 1 and 4" stuff, and from left, and from right?

Can you guys point me in the right direction, so I can try to learn it? what is the name of the topic?

If I disagree with you, it's because you're wrong.

In reply to cold pizza 2
11/18/11
cold pizza 2:

i didn't come from a private school or anything, so maybe that's why, but...

none of this shit seems to me like middle school or even high school level math.

what is this "i = 1 and 4" stuff, and from left, and from right?

Can you guys point me in the right direction, so I can try to learn it? what is the name of the topic?

The short answer: slope/intersect form. It appears that non-target is trying to solve it as a linear programming model (although is isnt in y=mx+b form yet), but I do not think this problem requires that degree of complication.

In reply to Im not very creative
11/18/11
Im not very creative:
cold pizza 2:

i didn't come from a private school or anything, so maybe that's why, but...

none of this shit seems to me like middle school or even high school level math.

what is this "i = 1 and 4" stuff, and from left, and from right?

Can you guys point me in the right direction, so I can try to learn it? what is the name of the topic?

The short answer: slope/intersect form. It appears that non-target is trying to solve it as a linear programming model (although is isnt in y=mx+b form yet), but I do not think this problem requires that degree of complication.

As I mentioned, the solution requires nothing more than algebra. Of course, algebra can mean anything from middle school to PhD level stuff, but a typical Algebra 2 course is sufficient here.

For what its worth, this problem came from ________________

Edit - will cite source after it is solved

11/18/11

Algebra 2 is what? "pre-calc" or something like Linear Algebra

If I disagree with you, it's because you're wrong.

In reply to nonTargetChimp
11/18/11
nonTargetChimp:

Multiply both sides by 2
u.(2k - u^2 + k) = v(2k - v^2 + k)
u.(3k - u^2) = v.(3k - v^2)
3k(u-v) - (u^3 - v^3) = 0

Case 1: u = v. Done

Case 2: 3k = u^2 + v^2 - uv
or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

u^3 - v^3 = (u-v)(u^2 + v^2 + uv), not (u-v)(u^2 + v^2 - uv),

11/18/11

pre-calc

though to be completely honest, to understand/complete the proof (and be able to show that it is true on a deep level), a mathematical proofs course would help

anyways i'll be in the chat room for a bit if anyone wants to discuss this problem

In reply to Dr Joe
11/18/11
Dr Joe:

or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)

No I'm not wrong let me clarify further
3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

3(a^2+b^2) = 2(a^2+b^2) + (c^2 + d^2)

(a + b)^2 + (c + d)^2 - (a+b)(c+d)
= a^2+b^2 + 2ab + c^2 + d^2 + 2cd - (a+b)(c+d)

Therefore you can cross a^2+b^2 + c^2 + d^2 from both sides. Please double check it for yourself

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

In reply to nonTargetChimp
11/18/11
nonTargetChimp:
Dr Joe:

or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)

No I'm not wrong let me clarify further
3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)

3(a^2+b^2) = 2(a^2+b^2) + (c^2 + d^2)

(a + b)^2 + (c + d)^2 - (a+b)(c+d)
= a^2+b^2 + 2ab + c^2 + d^2 + 2cd - (a+b)(c+d)

Therefore you can cross a^2+b^2 + c^2 + d^2 from both sides. Please double check it for yourself

yeah my bad you are right. damn checking these proofs is surprisingly difficult. what about the comment above?

In reply to is-t
11/18/11

is-t:

u^3 - v^3 = (u-v)(u^2 + v^2 + uv), not (u-v)(u^2 + v^2 - uv),

Oh yeah you're right


Goddamn it. No way I'm spending anymore of my Friday night time to do this.. Good luck fellow primates.

P/S: Why can't I edit my previous post anymore ????

Nobody wants to work for it anymore. There's no honor in taking the after school job at Mickey D's. Honor's in the dollar, kid.

11/18/11

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

Show that:
a+b=c+d

a,b,c,d >0
Pf: Assume a+b=c+d

(a+b)^2=(c+d)^2
a^2+2ab+b^2=c^2+2cd+d^2
a^2+b^2-c^2-d^2=2cd-2ab
0=2cd-2ab
0=cd-ab
ab=cd

(a+b)^3=(c+d)^3
a^3+3a^2b+3ab^2+b^3=c^3+3c^2d+3cd^2+d^3
a^3+b^3-c^3-d^3=3c^2d+3cd^2-3a^2b-3ab^2
0=3cd(c+d)-3ab(a+b)
3ab(a+b)=3cd(c+d)
ab(a+b)=cd(c+d)
Using shit above, ab=cd --> (a+b)=(c+d)

Closer?

11/18/11

Here is an idea.... (someone can do the algebra grunt work and get the credit)

Lets get rid of one variable first

rewriting first equation

a = (c^2 + d^2 - b^2) ^ (1/2)

rewriting second equation

a = (c^3 + d^3 - b^3) ^ (1/3)

now we can get rid of "a"

(c^2 + d^2 - b^2) ^ (1/2) = (c^3 + d^3 - b^3) ^ (1/3)

******NOW THIS IS WHERE IT GETS COMPLICATED **********

we can rewrite the equation above isolating "b" so it is in terms of "c" and "d"

i don't know how to factor that shit, but in theory we would have an equation like

b = ( nasty ass equation with "c" and "d" )
--------------
we could do the same steps up to this point, but instead of getting rid of "a" we get rid of "b"

b = (c^2 + d^2 - a^2) ^ (1/2)
b = (c^3 + d^3 - a^3) ^ (1/3)

then

(c^2 + d^2 - a^2) ^ (1/2) = (c^3 + d^3 - a^3) ^ (1/3)

again we isolate "a"

a = (nasty ass equation with "c" and "d")

******** now this is where everything should tie ************

this two equations should be exactly the same

b = ( nasty ass equation with "c" and "d" )
a = (nasty ass equation with "c" and "d")

therefore

a = b
------------------------------------------
then

c = d (by doing the dame process for "c" and "d" )

by substitution in first equation:

a^2+ a^2 = c^2 + c^2

2 a^2 = 2 c^2

simplified

a = c

and is done.

if we know that a = b , c = d , a = c

then

a + b = c + d

11/18/11

Here is a list of 42 methods of mathematical proofs.......might actually help

1. Proof by Obviousness: "The proof is so clear that it need not be mentioned."
2. Proof by General Agreement: "All in Favor?..."

3. Proof by Imagination: "Well, we'll pretend its true."
4. Proof by Convenience: "It would be very nice if it were true, so ..."

5. Proof by Necessity: "It had better be true or the whole structure of mathematics would crumble to the ground."
6. Proof by Plausibility: "It sounds good so it must be true."

7. Proof by Intimidation: "Don't be stupid, of course it's true."
8. Proof by Lack of Sufficient Time: "Because of the time constraint, I'll leave the proof to you."

9. Proof by Postponement: "The proof for this is so long and arduous, so it is given in the appendix."
10. Proof by Accident: "Hey, what have we here?"

11. Proof by Insignificance: "Who really cares anyway?"
12. Proof by Mumbo-Jumbo: " For any epsilon> 0 there exists a corresponding delta > 0 s.t. f(x)-L < epsilon whenever x-a < delta"

13. Proof by Profanity: (example omitted)
14. Proof by Definition: "We'll define it to be true."

15. Proof by Tautology: "It's true because it's true."
16. Proof by Plagiarism: "As we see on page 238 ..."

17. Proof by Lost Reference: "I know I saw this somewhere ..."
18. Proof by Calculus: "This proof requires calculus, so we'll skip it."

19. Proof by Terror: When intimidation fails ...
20. Proof by Lack of Interest: "Does anyone really want to see this?"

21. Proof by Illegibility: " Y= a D Th thae"
22. Proof by Logic: "If it is on the problem sheet, then it must be true."

23. Proof by Majority Rule: Only to be used if General Agreement is impossible.
24. Proof by Clever Variable Choice: "Let A be the number such that this proof works."

25. Proof by Tessellation: "This proof is just the same as the last."
26. Proof by Divine Word: "And the Lord said, 'Let it be true,' and it came to pass."

27. Proof by Stubbornness: "I don't care what you say! It is true!"
28. Proof by Simplification: "This proof reduces to the statement, 1 + 1 = 2."

29. Proof by Hasty Generalization: "Well, it works for 17, so it works for all reals."
30. Proof by Deception: "Now everyone turn their backs ..."

31. Proof by Supplication: "Oh please, let it be true."
32. Proof by Poor Analogy: "Well, it's just like ..."

33. Proof by Avoidance: Limit of Proof by Postponement as t approaches infinity.
34. Proof by Design: "If it's not true in today's math, invent a new system in which it is."

35. Proof by Intuition: "I just have this gut feeling ..."
36. Proof by Authority: "Well, Bill Gates says it's true, so it must be."

37. Proof by Vigorous Assertion: "And I REALLY MEAN THAT!"
38. Proof by A.F.K.T. Theorem: "Any Fool Knows That!"

39. Proof by vigorous handwaving: Works well in a classroom.
40. Proof by seduction: "Convince yourself that this is true!"

41. Proof by accumulated evidence: "Long and diligent search has not revealed a counterexample."
42. Proof by Divine Intervention: "Then a miracle occurs ..."

11/18/11

Ok wait, if this is as simple as I think it is, I will seriously be kicking myself for having wasted my Thursday and Friday evenings. Here goes.

Take (a+b) and (c+d). Square both sides. In this case you get:

a^2 + 2ab + b^2 and c^2 + 2cd + d^2. Now, we know that a^2 + b^2 = c^2 + d^2. Hence,

a^2 + 2ab + b^2 = a^2 + 2cd + b^2. Subtract a^2 + b^2 from both sides we have:

ab = cd.

Now, we know that a^3 + b^3 = c^3 + d^3. If we factor each side out,

(a+b)(a^2 - ab + b^2) = (c+d)(c^2 - cd + d^2).

Now again we know that a^2 + b^2 = c^2 + d^2, and we also know that ab = cd. Hence,

(a+b)(a^2 - ab + b^2) = (c+d)(a^2 - ab + b^2).

Therefore a+b = c+d.

Seriously? OMfg.

Just Do It

11/18/11

As far as I can see, the only condition on a,b,c and d is that

a^2 - ab + b^2 and c^2 - cd + d^2 are non-zero. Other than that, really there are no conditions, though I think its safe to keep them all positive real numbers (not necessarily integers). I think QED no? ...

Just Do It

In reply to maximumlikelihood
11/18/11
maximumlikelihood:

Ok wait, if this is as simple as I think it is, I will seriously be kicking myself for having wasted my Thursday and Friday evenings. Here goes.

Take (a+b) and (c+d). Square both sides. In this case you get:

a^2 + 2ab + b^2 and c^2 + 2cd + d^2. Now, we know that a^2 + b^2 = c^2 + d^2. Hence,

a^2 + 2ab + b^2 = a^2 + 2cd + b^2. Subtract a^2 + b^2 from both sides we have:

ab = cd.

Now, we know that a^3 + b^3 = c^3 + d^3. If we factor each side out,

(a+b)(a^2 - ab + b^2) = (c+d)(c^2 - cd + d^2).

Now again we know that a^2 + b^2 = c^2