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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.
Find the shortest solution you can; I have one that only takes 4 lines of working.
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#2
(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.
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.
Spoiler:
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Suppose the equation has a solution, and w.l.o.g. 
.
But
. The result follows.
.But
. The result follows.
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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.
One can get this pretty snappily with the contrapositive. Took me quite a while to notice this.
Spoiler:
Show
Suppose the equation has a solution, and w.l.o.g. .
But. The result follows.
.But
. The result follows.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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#4
(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...
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...
Moreover this question was initially set before the theorem was proved.
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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.
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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#6
(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.
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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(Original post by Zacken)
He's a well known troll, this is his ~10/11th account. I wouldn't bother replying to him.
He's a well known troll, this is his ~10/11th account. I wouldn't bother replying to him.
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