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#1 h Mar 24, 2025, 3:42 PM

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Let $p$ be a prime number and $x_1, x_2,...,x_p$ be integers satisfying $p \mid x_1^n+x_2^n+...+x_p^n, \forall n \in \mathbb{N}$ Prove that $p \mid x_1-x_2$ .

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#2 h Mar 24, 2025, 5:02 PM

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Let $P(x) = (x - x_1)(x - x_2) \dots (x - x_p) = x^p + a_{p-1}x^{p-1} + \dots + a_1x + a_0$ .

Then, $p \mid \sum x \Rightarrow p \mid a_{p-1}$

$p \mid \sum x$ , $p \mid \sum x^2 \Rightarrow p \mid \sum x_i x_j \Rightarrow p \mid a_{p-2}$

Similarly, by Newton's identities, $p \mid a_k$ for all from 1 to $p-1$

$\Rightarrow P(x) \equiv x^p + a_0 \equiv x + a_0 \mod p$

Also, $P(x_i) = 0$ for all $i$

$\Rightarrow x_i + a_0 \equiv 0 \Rightarrow x_1 \equiv x_2 \equiv -a_0$

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#3 h Mar 24, 2025, 5:04 PM

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Call $(x_1,\dots,x_p)$ nice if the conclusion holds. We prove that for any nice $(x_1,\dots,x_p)$ , the number of $x_i$ for which $p\mid x_i$ is either 0 or $p$ . To see this, let $k:=|i:x_i\equiv 0\pmod{p}|$ . Taking $n=p-1$ , we find by Fermat that $\textstyle \sum_{1\le j\le p}x_j^{p-1}\equiv p-k\equiv 0\pmod{p}$ , so $k\in\{0,p\}$ . Now, if $p\mid x_i,\forall i$ , we are done, else assume none of $x_i$ are divisible by $p$ .

Next, using binomial expansion, it is not hard to check that for any $k$ and any nice $(x_1,\dots,x_p)$ , $(x_1-k,\dots,x_p-k)$ is also nice. Taking $k=x_1$ , we must have $p\mid x_j-k$ for all $2\le j\le p$ , hence the conclusion.

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#4 h Mar 24, 2025, 5:31 PM

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already posted here

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