Number Theory or Algebra?

Claim: For any positive integer $n$ the function $p(n)$ will always be lesser than $n$ i.e. $p(n)<n$

Proof: let take an integer $n=\overline{a_na_{n-1}...a_1}$

Now $n=10^{n-1}a_n +10^{n-2}a_{n-1}+.....+a_1 > 10^{n-1}a_n >a_n a_{n-1} a_{n-2}....a_1$ because each $0 \leq a_i \leq 9$

So, the above claim is true

Here $p(n)=n^2-11n-23 <n \\ n^2-12n-23 <0 \\ (n-6)^2<23+36=59 \\ (n-6-\sqrt{59})(n-6+\sqrt{59})<0$

So, range of $n$ is $\sqrt{59}-6 < n < \sqrt{59}+6 \\ 1<n <14$

For $1<n<10$ the equation will be $n^2-11n-23=n \\ n^2-12-23=0$ .

Clearly it has no integral solution.

For $10 \leq n \leq 13$ easily we can check that $n=13$ is the only solution