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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 .

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

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

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