Am I proving the right thing? $Q : $ To find a divisor of $(a + b + c + d)^5 - (a^5 + b^5 + c^5 +d^5)$.
I generalized using multinomial coefficients that :
$(x_1 + x_2 + x_3 + ... + x_m)^n - ((x_1)^n + (x_2)^n + (x_3)^n + ... (x_m)^n)$  is always divisible by $n$.
Am I right here ?
As a result of this by putting $x_1 = 1, x_2 = 1..., x_m = 1$, I get $m^n - m$ has to be divisible by n. But that's not correct. A counter example is $(m = 2, n = 4)$.
Help me to correctly find the true answer to the generalized formula.
 A: You have the right idea, but it only necessarily works in general for primes. By Fermat's little theorem, you have for any integer $n$ and prime $p$ that
$$n^p \equiv n \pmod{p} \tag{1}\label{eq1A}$$
Thus, your expression gives
$$\begin{equation}\begin{aligned}
& (a + b + c + d)^5 - (a^5 + b^5 + c^5 +d^5) \\
& \equiv (a + b + c + d) - (a + b + c + d) \\
& \equiv 0 \pmod{5}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
i.e., $5$ is always a divisor.
A: To complement the other answer: there are two other guaranteed divisors, $2$ and $3$.
Observe that mod $3$, we have $0^5 = 0, 1^5 = 1, 2^5 = 2$ and so it is similarly the case that
$$n^5 = n \mod 3$$
and hence your expression is $0$ mod $3$. The same works for $2$.
The above argument works only for $2,3,5$. For take a prime $q$ and suppose our congruence holds,
$$n^5 = n \mod q$$
for all $n \in \mathbb Z/q\mathbb Z$. This is a polynomial of degree $5$ and so it has at most $5$ different roots. But $\mathbb Z/q\mathbb Z$ has $q$ elements which are all roots of the above, and so $q\leq 5$ and we have checked that all those cases work.
Also, by considering the examples $1,1,1,1$ and $1,1,2,3$ for $a,b,c,d$ you can see that there are no other guaranteed divisors.
