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Recently I carried out an elementary proof of the following assertion, which is a special case of Fermat's last theorem:

If $p$ is an odd prime and $x, y, z > 0$ are integers such that $(x, y) = 1$ and $z-y \mid x$ and $z-y \neq 1,$ then $x^{p} + y^{p} \neq z^{p}$

I am wondering, has there already been any proof in literature?

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  • $\begingroup$ Forgot to say: I have found not $\endgroup$ – Megadeth Jul 1 '14 at 14:07
  • $\begingroup$ Does your proof actually use the hypothesis that $p$ is odd? $\endgroup$ – Gerry Myerson Jul 22 '14 at 7:31
  • $\begingroup$ @Gerry Thank you for your interest. Yes, the proof does. Moreover, the proposition above partitions Fermat's last theorem into exactly three cases: $z-y \mid x$ and $z-y \neq 1$, $z-y=1$, and $z-y \nmid x.$ But a casewise proof is, so to speak, not elegant ~ $\endgroup$ – Megadeth Jul 22 '14 at 7:46
  • $\begingroup$ Here is a possible partial elementary agrument. if $C^{2 n}- B^{2 n} = A^{2 n}$ for n >= 2 , C being an odd integer and B an even integer then (C^n -B^n)(C^n + B^n) = $A^{2 n}$. Since C is odd and B is even then (C^n -B^n) and (C^n + B^n) are coprime therefore (C^n - B^n) is a integer to the exponent (2 n) and so is (C^n + B^n). Let $h^{2 n}$ +B^n = C^n and $k^{2 n}$ - B^n = C^n so $h^{2 n} + k^{2 n}$ = (2 (C^n)) ; h and k being odd so $(C^n)\equiv 1\pmod{4}$. Therefore if $A^{2 n} + B^{2 n} = C ^{2 n}$ and C^n is not congruent to 1 mod 4 ( C odd and B even) then Contradiction. $\endgroup$ – user128932 Sep 16 '14 at 18:00
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This result is an immediate consequence of the "Relations of Barlow". Without loss of generality we can assume that $p$ is not a factor of $x$. Then these relations state that there are coprime integers $r$ and $s$ such that $$x=rs,z-y=s^p$$ Your condition that $z-y$ divides $x$ would mean that $s^p$ divides $rs$ and is clearly impossible.

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