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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

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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. .

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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Browse my blog as a WSO contributing author

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.

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