Find a positive integer $n$ such that its digits product is $n^2 -13n-25$
If we assume that $n$ is a two digit integer then let $n=10k+r$ where $1 \leq k \leq 9 $ and $0 \leq r \leq 9$ . Plugging $n=10k+r$ in $n^2 -13n-25$ we get $kr=(10k+r)^2 -13(10k+r)-25$. Manipulating this expression as a quadratic for $r$ we conclude that $n=15$ is a solution.
I can't generalize it for any $n$ greater than two digit. Can anyone help? Thanks in advance.