Find primes satisfying this Find all prime numbers n such that 
$$n | 6^{n}(n-4)! + 10^{3n}$$
Can't seem to figure out how to start this, any hint would be helpful.
Will Fermat's or Wilson's theorem be used?
 A: Indeed, both Wilson’s theorem and Fermat’s little theorem are relevant. If $n\gt3$, then since $n$ is prime $(n-3)(n-2)(n-1)$ is coprime with $n$, so the desired divisibility is equivalent to
$$
n\mid\left(6^n(n-4)!+10^{3n}\right)(n-3)(n-2)(n-1)=6^n(n-1)!+10^{3n}(n-3)(n-2)(n-1)\;.
$$
Now you can use Wilson’s theorem $(n-1)!\equiv-1\bmod n$ and Fermat’s little theorem $a^n\equiv a\bmod n$ to simplify this. Since you only asked for a hint, I’ll leave the rest to you.
A: Caution: Not a hint, but a complete solution.
If $n\in \{2,3\}$ then $(n-4)!$ doesn't exist.
If $n=5$ then $6^n(n-4)!+10^{3n}\equiv 1^n(1)!+10^{15}\equiv 1 \not \equiv 0\mod 5.$
If $n$ is prime then (i) by Fermat's Little Theorem, $6^n\equiv 6 \mod n$ and $10^{3n}=1000^n\equiv 1000\mod n,$ and  (ii)  by Wilson's Theorem, $(n-1)!\equiv -1\mod n .$ And we have $-6 \equiv (n-3)(n-2)(n-1) \mod n.$
So if $n\ge 7$ is prime then modulo $n$ we have $$6^n(n-4)!+10^{3n}\equiv 0\iff$$ $$ 6(n-4)!+1000\equiv 0 \iff$$ $$ 6(n-4)!(-6)+1000(-6)\equiv 0 \iff$$ $$ 6(n-4)!(n-3)(n-2)(n-1)+1000(-6)\equiv 0 \iff$$ $$ 6(n-1)!+1000(-6)\equiv 0\iff$$ $$ 6(-1)+1000(-6)\equiv 0\iff$$ $$ (-1)+1000(-1)\equiv 0 \iff$$ $$ n\in \{7,11,13\}.$$
