## Pages

So it seems that there are some math people around. To follow up on //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)

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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No such luck.

http://www.wolframalpha.com/input/?i=%28a^3+%2B+b^3%29%2F%28a^2+%2B+b^2%29

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Fail.

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

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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.

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I just realized how stupid my answer was.

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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?

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

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Take the square root of each variable?

I eat success for breakfast...with skim milk

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

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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

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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.

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

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I fail as well.

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

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SB for the chuckle

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Math Proofs....

HUMBUG!!!Get busy living

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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

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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)

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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.

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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);

}

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this doesn't compile. weird.

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this doesn't compile. weird.[/quote]

Wtf are you trying to do, showing off that you can program at 101 level ?

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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Wtf are you trying to do, showing off that you can program at 101 level ?[/quote]

take it easy and stop being so uptight. it was just me fucking around. go out and have a beer. you seem stressed out from life.

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LOL you tried compiling it? i didnt even say which language.

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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

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no it doesnt.

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

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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

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

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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.

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Please embarrass everyone by doing this right now.

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I give up do it

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

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You have only had 6 pages of examples (right or wrong) to help you. Please bestow your genius upon me.

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I'm with Will I'm sure he did it the way I did it. Many people got burned for the sheer fact you need more than just regular algebra to make a simple proof.

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Word. A solution does not equal a proof. You can easily manipulate # into your said proof to make it hold true, which is where myself went wrong. I need to look over WH proof TBH.

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I am not with Will on this one.

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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

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

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Lol hoping you overlooked that

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

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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

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yeah i realized that too late lol. fail.

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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

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

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

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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

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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.

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.

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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

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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.

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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

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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.

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

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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.

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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

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

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

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No

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Dr. Joe you're such a tease.

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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

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

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

Where I unload on Twits and take verbal S***s

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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?

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Not at all. Sorry.

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Apparently the answer is no.

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...

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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

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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

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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

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

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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)

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

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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.

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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.

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

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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?

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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!

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I was wondering this same thing

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

Get busy living

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

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

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

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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 ;)

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...

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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

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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.

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

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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

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

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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.

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This gave me a headache.

Most people do things to add days to their life. I do things to add life to my days.Browse my blog as a WSO contributing author## Want to Vote on this Content?! No WSO Credits ?

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.

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

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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)

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

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

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But the problem remains open...

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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.

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

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

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here's a proof:

http://en.wikipedia.org/wiki/Pythagorean_theorem

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

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Proof that you're retarded

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lol

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

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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.

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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 ?

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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.

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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

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nice!

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You forgot to change the sign in the following step:

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missed that...

still waiting on a proof!

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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 !?!?!?!?!?!?!?!

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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.

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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.

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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.

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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 ?

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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...

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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.

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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.

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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.

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

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

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^very nice

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

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

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

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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

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Not limited to integers

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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.

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I guess n

nn-n*n = kturns to

n = k/n*n +1

so just one solution.

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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.

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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.

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^ wow that was retarded sorry forget that

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

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

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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.

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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.

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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.

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dude

?

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

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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

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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.

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^^^

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

Get busy living

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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.

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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

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He's like Yoda or something.

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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.

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I got something short and elegant for you.

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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..

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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.

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

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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.

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

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You're fucking with us right Dr. Joe?

Just Do It

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Given A = B = C = D = n

And n = 1

Is true

What about n =/= 1?

Get busy living

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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.

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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.