If 25 divides $a^5 + b^5 + c^5 + d^5 + e^5$ then prove that 5 divides $abcde$. I solved the following problem and giving the outline of my solution. I'm curious to know if a shorter and more elegant solution exists.
Problem
Let $a,b,c,d,e \in \mathbb{N}$ such that $25 | (a^5 + b^5 + c^5 + d^5 + e^5)$. Prove that $5 | abcde$.
Solution Outline
Use contradiction.
$5 \nmid abcde \implies 5 \nmid a,b,c,d,e$.
If $n = 5q + r$ then $n^5 = 25k + r^5$ which can be proved by Binomial Expansion. Thus $n \equiv r (mod \; 5) \implies n^5 \equiv r^5 (mod \; 25)$.
Now possible values of $r$ are $1,2,3,4$. Hence possible values of $r^5 (mod \; 25)$ are $1,7,-7,-1$. (We assume that the set of remainders modulo 25 is $\{-12, -11, \dots, -1, 0, 1, \dots, 11, 12 \}$). Thus $a^5, b^5, c^5, d^5, e^5 \equiv 1,7,-7,-1 (mod \; 25)$ in some order.
Then I spent a major chunk of my solution proving that $x_1 + x_2 + x_3 + x_4 + x_5 = 0 (mod \; 25)$ has no solutions if $x_1, x_2, x_3, x_4, x_5 \in \{1, 7, -7, -1 \}$. This part of the proof was essentially a brute force attack, and I'm not fully satisfied with it.
Question
I'm curious to know if this last part has a shorter/more elegant solution. Also, I'd be interested in any other solution overall.
Thanks!!

Since $5^1\binom{5}{1}=25$, we have $(n+5)^5\equiv n^5\pmod{25}$. Thus, we can generate the table
$$
\begin{array}{c|c}
n\pmod{5}&n^5\pmod{25}\\
0&0\\
1&1\\
2&7\\
3&-7\\
4&-1\\
\end{array}
$$
Since there are no solutions to $7x+y=25$ with $|x|+|y|\le5$, we must be looking for $7x+y=0$ with $|x|+|y|\le5$, which has only one solution: $x=y=0$. This corresponds to an equal number of $7\pmod{25}$ and $-7\pmod{25}$ and an equal number of $1\pmod{25}$ and $-1\pmod{25}$. Since $5$ is odd, there must be at least one $0\pmod{25}$ among $a^5,b^5,c^5,d^5,e^5$. This means at least one $0\pmod{5}$ among $a,b,c,d,e$.
