Find all natural numbers $k \geq 4$ such that $ k|2(k-3)! +36 $.

Question

Find all natural numbers $k \geq 4$ such that, $$ k|2(k-3)! +36\,. $$

Answer 1

Either $k$ is prime or $k$ is composite.

If $k$ is composite

Then $k=4$ fails by trial and for all other possibilities ($k\ge6$) the factorial $(k-3)!$ includes factors of both $d$ and $k/d$ for some divisor $d$ of $k$. Thus $k\ge 6$ and $k$ composite will satisfy the divisibility requirement iff $k|36$ admitting the solutions $k\in\{6,9,12,18,36\}$.

If $k$ is prime

Wilson's Theorem forces $(k-1)!\equiv -1\bmod k$. Divide by $(k-1)(k-2)\equiv 2$ and multiply by $2$ to render

$2(k-3)!\equiv -1\bmod k$

This means the right hand side of the divisibility relation is $\equiv 35$ and prime solutions for $k$ must be factors of that number. Thus prime values of $k$ are $\in\{5,7\}$.

The complete solution set is therefore $\{5,6,7,9,12,18,36\}$.