Knowing that $a_m = 2a_{m - 1}(a_{m - 1} + 1)$, find $n$ such that $a_1^2 + a_2^2 + \cdots + a_{m - 1}^2 + a_m^2 + n^2$ is a square number.

In general, this is not possible: Assume there are $b_m\ge0$ with $$a_1^2+\cdots +a_m^2+n^2=b_m^2.$$ For example, if $a_0=1$, we find $a_1=4$, so $$b_1^2=16+n^2.$$ Then $16=b_1^2-n^2=(b_1+n)(b_1-n)$, where both factirs have same, hence even, parity. This allows only $b_1+n=8$, $b_1-n=2$, so necessarily $n=3$, $b_1=5$. But $a_2=2a_1(a_1+1)=40$, and we'd need $$b_2^2=a_1^2+a_2^2+n^2=1625,$$ which is not a square.