Let $x_1, x_2, \cdots, x_n$ be integers such that

$$5040 \mid x_1+x_2+\cdots+x_n,\\ 5040 \mid x_1^3+x_2^3+\cdots+x_n^3,\\ 5040 \mid x_1^5+x_2^5+\cdots+x_n^5, $$

Then prove that $$5040 \mid x_1^7+x_2^7+\cdots+x_n^7.$$

I don't even know where to start... All I know is that $5040$ is a highly composite number with $60$ divisors and that it has all consecutive primes $2$, $3$, $5$ and $7$ in its factorization.


Note that $5040=7!$. Let $f(x) = (x-3)(x-2)(x-1)x(x+1)(x+2)(x+3)\in \mathbb{Z}[x]$, which is an odd polynomial and so $f(x) = x^{7}+ax^{5}+bx^{3}+cx$ for some $a, b, c\in \mathbb{Z}$. Then for any $n\in \mathbb{Z}$, $f(n)$ is a multiple of $7!$ since it is a product of seven consecutive integers. Now if $\sum_{i}x_{i}^{m}$ are multiple of $7!$ for $m=1, 3, 5$, then $$\sum_{i}x_{i}^{7} = \sum_{i}f(x_{i}) -a\sum_{i}x_{i}^{5}-b\sum_{i}x_{i}^{3}-c\sum_{i}x_{i}$$ is also multiple of $7!$.

  • 3
    $\begingroup$ Nice!${}{}{}{}$ $\endgroup$ – quasi Feb 21 '18 at 10:33

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