Sum and sum to the seventh power

Assume that a set of integers $x_1, x_2, x_3, ... x_n$ can be chosen such that $x_1 + x_2 + x_3 + ... + x_n = 1280$ and $x_1^7 + x_2^7 + x_3^7 + ... + x_n^7 = 2018$ .

Since by Fermat's Little Theorem $x^7 \equiv x \pmod 7$ , $x_1^7 + x_2^7 + x_3^7 + ... + x_n^7 \equiv x_1 + x_2 + x_3 + ... + x_n \pmod 7$ , which implies that $1280 \equiv 2018 \pmod 7$ .

However, $1280 \equiv 6 \pmod 7$ and $2018 \equiv 2 \pmod 7$ , so $1280 \not\equiv 2018 \pmod 7$ , so we reject our assumption and conclude that there are no set of integers that can fulfill the given conditions.

3^7 = 2187 > 2018, so the integers must consist of 1's and 2's. Let there be n 1's and m 2's. then n + 2m = 1280, and n (1^7) + m (2^7) = 2018. Since 2^7 = 128, the last equation may be written as: n + 128m = 2018. Substituting for n from the first equation, 1280 - 2m + 128m = 2018, or 126m = 738. But 738 is not divisible by 128. Therefore impossible. Ed Gray