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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.

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## 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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I just realized how stupid my answer was.

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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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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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Math Proofs....

HUMBUG!!!Get busy living

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

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Please embarrass everyone by doing this right now.

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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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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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Lol hoping you overlooked that

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I give up do it

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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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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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yeah i realized that too late lol. fail.

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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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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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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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.

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

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

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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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