Term $\times$ Term $=$ Term?

We note that $a_m = m^2 + 1$ , where $m$ is a positive integer. Let $k=97$ ; then:

$\begin{aligned} n^2 + 1 & = \left(k^2 +1 \right) \left((k+1)^2 + 1 \right) \\ & = k^2(k+1)^2 + k^2 + (k+1)^2 + 1 \\ \implies n^2 & = k^2(k+1)^2 + k^2 + (k+1)^2 \\ & = k^4 + 2k^3 + 3k^2 + 2k + 1 \\ & = (k^2+k+1)^2 \\ \implies n & = k^2 + k + 1 = 97^2 + 97 + 1 = \boxed{9507} \end{aligned}$

Clearly $a_n = n^2+1,$ and $(n^2+1)((n+1)^2+1) = n^4+2n^3+3n^2+2n+2 = (n^2+n+1)^2+1,$ so $a_n a_{n+1} = a_{n^2+n+1}.$

Plugging in $n=97$ gives $n^2+n+1 = \fbox{9507}.$