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Can we reduce Fermat last theorem problem to the case $z=x+1$ where $x^n + y^n = z^n$?

Why am I asking that? I found in that case and in case $n=3$ that difference of cubes: $1$,$7$,$19$,$37$ http://oeis.org/A003215

is a combination of diferrence of next sequence: http://oeis.org/A011934

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Prior to Wiles's theorem that there is no integer solution, there was no such reduction, and Wiles's proof does not go through such a reduction. –  André Nicolas Apr 8 '12 at 20:17
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I think we may be able to reduce to the cases $n=1$ or $n=2$. The condition specified adds nothing in either of these cases. –  Mark Bennet Apr 8 '12 at 20:20
    
I don't get what you mean: Centered hexagonal numbers are combinations of differences of $\sum (-1)^{n+1} n^3$? –  draks ... Apr 8 '12 at 20:33
    
1-0=1,7-0=7,20-1=19,44-7=37,81-20=61,135-44=91 and so on.... –  Bojan Vasiljević Apr 8 '12 at 20:43
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What has long been known (and proved by elementary methods) , in the crucial case that $n$ is prime and $x,y$ and $z$ are pairwise coprime, is that in the case ${\rm gcd}(n,xyz) =1,$ we have $z-x = u^{n}, z-y = v^{n},$ and $x+y = w^{n},$ for integers $u,v$ and $w.$ In the case that $n$ is prime and $n$ divide $xyz$ (but $x,y$ and $z$ still pairwise relatively prime) we either have $z-x = n^{n-1}u^{n}$ or $z-x = u^{n}$ for some integer $u.$

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