If 25 divides $a^5 + b^5 + c^5 + d^5 + e^5$ then prove that 5 divides $abcde$.

$x_1 + x_2 + x_3 + x_4 + x_5 = 0 \pmod{25}\,$ has no solutions if $x_1, x_2, x_3, x_4, x_5 \in \{1, 7, -7, -1 \}$

The sum lies between $5 \cdot(-7) = -35$ and $5 \cdot 7 = 35\,$. The sum of $5$ odd numbers cannot equal an even, which excludes $0\,$, and the only other multiples of $25$ in that range are $\pm 25\,$. By symmetry, it is enough to consider the case where the sum equals $25\,$, and that can be easily excluded by noting that it would require $3$ or $4$ of the values to be $7$, but neither such combination gives a solution.