How do I prove this claim? Claim :Let $p$ be a prime and $m \geq 2$ be an integer. Prove that the equation $ \frac{ x^p + y^p } 2 = \left( \frac{ x+y } 2 \right)^m $ has a positive integer solution $(x, y) \neq (1, 1)$ if and only if $m = p$.
Thank you for your help .
 A: $ p > 2$  $ \implies$  $ (x + y)|x^{p} + y^{p}$
$ q > 2$ ,$ q$ is prime,$ q^{a}\parallel{}x,q^{b}\parallel{}y$ ,$ a\leqq$
$ \implies$  $ ap = am$  $ \implies$  $ p = m$
$ \implies$  $ (x,y) = 1$  or $ (x,y) = 2^{\alpha}$
$ (x,y) = 2^{\alpha}$,$ x = 2^{\alpha}.x_{1},y = 2^{\beta}.y_{1}$ ,$ {\beta}\geq {\alpha}$
$ x + y = 2^{\alpha}[x_{1} + y_{1}.2^{{\beta} - {\alpha}}]$
$ r$  is prime, $ r|[x_{1} + y_{1}.2^{{\beta} - {\alpha}}]$  $ \implies$  $ (r,x) = (r,y) = 1$
because if $ (r,y) > 1$  $ \implies$  $ (r,x) > 1$
and $ r|[x_{1} + y_{1}.2^{{\beta} - {\alpha}}]$  if $ {\beta} > {\alpha}$  $ \implies$   $ r > 2$
and $ r|x_{1},r|y_{1}$  Contradiction!
$ \implies$  $ {\alpha} = {\beta}$
$ \implies$  $ x = 2^{\alpha}.x_{1},y = 2^{\alpha}.y_{1}$
$ r|x_{1} + y_{1}$,  if $ r > 2$  $ \implies$  $ r\nmid x,r\nmid y$
$ (r,x) = (r,y) = 1$  $ \implies$  $ v_{r}(x^{p} + y^{p}) = v_{r}(x + y) + v_{r}(p) = m.v_{r}(x + y)$
if $ p\not = r$  $ \implies$  $ m = 1$  Contradiction
if $ p = r$  $ \implies$  $ m = 2$
$ \implies$  $ x^{p} + y^{p} = \frac {(x + y)^{2}}{2}$  $ \implies$  $ p\leq 2$  $ \implies$  $ p = 2$ for $ x = y$  
$ \implies$  $ p = m = 2$  
if $ x^{p} + y^{p} = 2^{c}$,and $ v_{2}(x) = v_{2}(y) = {\alpha}$
$ \implies$  $ x_{1}^{p} + y_{1}^{p} = 2^{s}$
$ x_{1} = 2^{k} - y_{1}$
$ (2^{k} - y_{1})^{p} + y_{1}^{p} = 2^{s}$
$ 2^{kp} - 2^{k(p - 1)}.y_{1} + ... + 2^{k}y_{1}^{p - 1} = 2^{s}$
$ \implies$  $ 2^{k}\parallel{}2^{s}$  $ \implies$  $ k = s$
$ \implies$  $ (2^{s} - y_{1})^{p} + y_{1}^{p} = 2^{s}$
$ 2^{sp} - 2^{s(p - 1)}.y_{1} + .. + 2^{s}y_{1}^{p - 1} = 2^{s}$
$ \implies$ $ 2^{sp}\geq2^{s(p - 1)}.y_{1}$
$ \implies$  $ 2^{s}\geq2^{s}.y_{1}^{p - 1}$
$ \implies$  $ y_{1} = 1$
if $ (x,y) = 1$  $ \implies$,$ r$ is prime $ r|x + y$  and $ (r,x) = (r,y) = 1$
the rest is directly tricky lemma  .
