How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem

$$n^5+29n-(n-2)(n-1)n(n+1)(n+2)=5n^3+25n=5(n-1)n(n+1)+30n$$
$$\implies n^5+29n=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{The product of }5\text{ consecutive integers}}+5\underbrace{((n-1)n(n+1))}_{\text{The product of }3\text{ consecutive integers}}+30n$$
See The product of n consecutive integers is divisible by n factorial
