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I am wondering about correctness of the following theorem. It looks very simple to me but I can't seem to find a mistake. Hopefully anyone will take a look at it.

Theorem:

Let:

$\quad\quad p$ an odd prime,

$\quad\quad \gcd(x,y,z) = 1,$

$\quad\quad x^p = y^p + z^p,$

Then:

$\quad\quad p \equiv 1 \pmod{r^p}$ for some $r$

Proof:

$\quad\quad$ Assume $p \not | z$ and consider the following congruence:

$\quad\quad\quad\quad (x^p - y^p)/(x - y) \equiv px^{p - 1} \equiv py^{p - 1} \pmod{x - y}$

$\quad\quad$ In which division has to be done before the modulo-operation.

$\quad\quad$ It follows that:

$\quad\quad\quad\quad \gcd(x - y,(x^p - y^p)/(x - y))$

$\quad\quad\quad\quad = \gcd(x - y,px^{p - 1})$

$\quad\quad\quad\quad = \gcd(x - y,p) = 1$

$\quad\quad$ From which we conclude that:

$\quad\quad\quad\quad x - y = r^p,$

$\quad\quad\quad\quad (x^p - y^p)/(x - y) = s^p$

$\quad\quad$ for some $r,s$ with $\gcd(r,s) = 1, z = rs$

$\quad\quad \implies s^p - px^{p - 1} \equiv (x^p - y^p)/(x - y) - px^{p - 1} \equiv 0 \pmod{x - y}$

$\quad\quad$ Because:

$\quad\quad\quad\quad s^p \equiv px^{p - 1} \equiv py^{p - 1} \pmod{x - y}$

$\quad\quad x,y$ are interchangeable

$\quad\quad \implies x^p - y^p - (x - y)px^{p - 1} \equiv 0 \pmod{(x - y)^2}$

$\quad\quad \implies (x^p - px^p) - (y^p - py^p) \equiv 0 \pmod{(x - y)^2}$

$\quad\quad \implies (1 - p)(x^p - y^p) \equiv 0 \pmod{(x - y)^2}$

$\quad\quad \implies p \equiv 1 \pmod{x - y}$

$\quad\quad \implies p \equiv 1 \pmod{r^p}$

  • $\quad\quad p \equiv 1 \pmod{r^p}$ if $r$ is an integer then $r^p$>p. I am only an engineer so may have got this wrong, but does this not suggest $p=1$ and not a prime as you suggest. – James Arathoon Nov 19 '17 at 00:48
  • Yes the idea is that it ends up on something impossible –  Nov 19 '17 at 00:50
  • ok but I'm still confused as then you say "consider the following congruence $(x^p - y^p)/(x - y) \equiv px^{p - 1} \pmod{x - y}$ " of the form $n \equiv b \pmod{a}$. As $n>a$ I am expecting $a>b$ that is $(x - y)>px^{p - 1}$ which is definitely not the case. Can you explain? – James Arathoon Nov 19 '17 at 01:38
  • Well, important is that division has to happen before the modulus-operation, forgot to mention that but it can be proven inductively for sure –  Nov 19 '17 at 01:45
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    A much simpler proof is, Take $r=1$. – Gerry Myerson Nov 19 '17 at 01:55
  • Yes, ok. Further inspection shows that $x,y$ are not interchangeable modulo $(x - y)^2$ so this won't work anyway. –  Nov 19 '17 at 02:27

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