# 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 .

• For 'only if', see here for a proof based on Lifting The Exponent Lemma (LTE). For 'if', take $(x,y)=(2,2)$. – user26486 Jun 5 '15 at 16:20

$p > 2$ $\implies$ $(x + y)|x^{p} + y^{p}$

$q > 2$ ,$q$ 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 .

• thank you for your answer , now it's seems good and clear to me – zeraoulia rafik Jun 5 '15 at 16:07