Example 6 (1989 Singapore Secondary School Mathematics Competition) Let $p_{n}=1^{n}+2^{n}+3^{n}+4^{n}$, find all positive integers $n$ that make $p_{n}$ divisible by 5.
Since $2 \equiv 2(\bmod 5), 2^{2} \equiv 4(\bmod 5), 2^{3} \equiv 3(\bmod 5), 2^{4} \equiv 1(\bmod 5)$, we have $2^{n} \equiv$ $2^{n+4}(\bmod 5)$. Similarly, $3^{n+4}=81 \cdot 2^{n} \equiv 2^{n}(\bmod 5), 4^{n+4} \equiv 256 \cdot 4^{n} \equiv 4^{n}(\bmod 5)$, thus $p_{n} \equiv$ $p_{n+4}(\bmod 5)$. And $p_{1} \equiv 0(\bmod 5), p_{2} \equiv 0(\bmod 5), p_{3} \equiv 0(\bmod 5), p_{4} \equiv 4(\bmod 5)$.
Therefore, for all positive integers $n$ that are not multiples of 4, $p_{n}$ is a multiple of 5.