$1! + 2! + ... + n! = k^3 + 6$ , find positive integers $n$ and $k$ satisfying the above equation.
Here I have approached this problem for $n \lt 5$ and got one of the solutions as $n = 4 , k = 3$, but for $n \gt 5$ all I can conclude is , that $k$ should be $3$ modulo $30$ , but this is not helping in finding integer solutions for $n \gt 5$ . Please suggest an approach.
From the well-known properties, we can prove that $x^3$ can give remainders $0, 1, 6$ in mod $7$. Therefore, $k^3 + 6$ will yield $6, 5, $ or $0$ in mod $7$. Checking the small cases, you will find that $n$ can't be greater than $7$ due to the inapplicability of $1! + ... + 6!$ for the mentioned remainders. Small cases can be checked easily.
Update: Although mod $7$ could give identity relations when $n \geq 7$ is considered, other users have suggested considering mod $13$ since it holds almost the same properties with $7$ only with more base cases and proving the inapplicability of $n \geq 13$ via obtaining $1! + 2! + ... + 12!$ different from above values.