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HapaxOromenon2
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Prove that the equation x^n + y^n = z^n, where n is an integer > 1, has no solution in integers x, y, and z, with 0 < x <= n, 0 < y <= n.

Find the shortest solution you can; I have one that only takes 4 lines of working.
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joostan
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(Original post by HapaxOromenon2)
Prove that the equation x^n + y^n = z^n, where n is an integer > 1, has no solution in integers x, y, and z, with 0 < x <= n, 0 < y <= n.

Find the shortest solution you can; I have one that only takes 4 lines of working.
One can get this pretty snappily with the contrapositive. Took me quite a while to notice this.
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Suppose the equation has a solution, and w.l.o.g. x \leq y &lt; z.
x^n=z^n-y^n=(z-y)\displaystyle\sum_{k=0}^{n-1}z^{n-1-k}y^k &gt; (z-y)ny^{n-1} \geq nx^{n-1}.
But x^n &gt; nx^{n-1} \iff x&gt;n. The result follows.
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HapaxOromenon2
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(Original post by joostan)
One can get this pretty snappily with the contrapositive. Took me quite a while to notice this.
Spoiler:
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Suppose the equation has a solution, and w.l.o.g. x \leq y &lt; z.
x^n=z^n-y^n=(z-y)\displaystyle\sum_{k=0}^{n-1}z^{n-1-k}y^k &gt; (z-y)ny^{n-1} \geq nx^{n-1}.
But x^n &gt; nx^{n-1} \iff x&gt;n. The result follows.
What I did: By Fermat's Last Theorem there are no solutions for n>2, so since we are given n>1, we need only consider n = 2. Thus we have that x^2 + y^2 = z^2 and 0<x<=2, 0<y<=2, so x and y can each only be 1 or 2.
Thus we check 1^2 + 1^2 = 2, 1^2 + 2^2 = 2^2 + 1^2 = 5, 2^2 + 2^2 = 8, none of which are square numbers. Thus the result follows.

Much simpler than your approach...
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joostan
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(Original post by HapaxOromenon2)
What I did: By Fermat's Last Theorem there are no solutions for n>2, so since we are given n>1, we need only consider n = 2. Thus we have that x^2 + y^2 = z^2 and 0<x<=2, 0<y<=2, so x and y can each only be 1 or 2.
Thus we check 1^2 + 1^2 = 2, 1^2 + 2^2 = 2^2 + 1^2 = 5, 2^2 + 2^2 = 8, none of which are square numbers. Thus the result follows.

Much simpler than your approach...
I disagree, assuming Fermat's Last Theorem is gigantic, the proof of it - monumental.
Moreover this question was initially set before the theorem was proved.
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HapaxOromenon2
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(Original post by joostan)
I disagree, assuming Fermat's Last Theorem is gigantic, the proof of it - monumental.
Moreover this question was initially set before the theorem was proved.
True. But it's still easier to understand than the thing you did, because you don't have to understand the proof of a theorem in order to understand it's statement and to use it to solve problems.
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Zacken
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(Original post by joostan)
I disagree, assuming Fermat's Last Theorem is gigantic, the proof of it - monumental.
Moreover this question was initially set before the theorem was proved.
He's a well known troll, this is his ~10/11th account. I wouldn't bother replying to him.
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HapaxOromenon2
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(Original post by Zacken)
He's a well known troll, this is his ~10/11th account. I wouldn't bother replying to him.
I demand that you cease and desist from defamation of me at once.
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