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Let $ p\equiv2\pmod 3$ be a prime number. Prove that the equation $x^p_{1}+x^p_{2}+\cdots+x^p_{n}+1=(x_{1}+x_{2}+\cdots+x_{n})^2$ has no integer solutions.

This problem is from the (Problems from the book) chapter 18 Quadratic reciprocity. Because this book problems have no answer. so How to use this methods to solve it?

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Modulo $2$, the LHS is congruent to $x_1 + \cdots + x_n + 1$ while the RHS is congruent to $x_1 + \cdots + x_n$.

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  • $\begingroup$ To make sure I'm understanding, this answer does not use the fact that $p\equiv2\pmod 3$, right? $\endgroup$ – Jason DeVito Mar 26 at 15:16
  • $\begingroup$ Yes, that's right. $\endgroup$ – punctured dusk Mar 26 at 15:17
  • $\begingroup$ Very nice! +1 from me. $\endgroup$ – Jason DeVito Mar 26 at 15:17
  • $\begingroup$ Very nice answer! $\endgroup$ – Mostafa Ayaz Mar 26 at 15:24
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Notice that:

$$(x_1+...+x_n)^2=x_1^2+...+x_n^2-2\prod_{i,j=1;i>j}^{n}x_ix_j$$

So:

$$1+x_1^p+...+x_n^p\equiv (x_1+...+x_n)^2 \pmod{2} $$$$ \Rightarrow 1+x_1^p+...+x_n^p\equiv x_1^2+...+x_n^2 \pmod{2} $$

Notice that $ \text{ord}(2)=1 $ so $x^y\equiv x \pmod{2}$ then:

$$ 1+x_1+...+x_n\equiv x_1+...+x_n \pmod{2} $$

$$ 1\equiv 0 \pmod{2} $$

:)

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