Question

Find all positive integers $n$ such that $3^{2n}+3n^2+7$ is a perfect square.

Answer 1

If $3^{2n}+3n^2+7$ is a perfect square

then, $3^{2n}+3n^2+7=b^2$ for some natural number b.

$⟹b^2>3^{2n}$

$⟹b>3^n$
$⟹b\ge 3^n+1$.
Thus $3^{2n}+3n^2+7=b^2\ge {(3^n+1)}^2=3^{2n}+{2.3}^n+1$.
$⟹2\cdot 3^n\le 3n^2+6$ ------(1)

if $n\ge 3,$ (1) cannot hold. One can prove this eithe by induction or by direct argument.
If $n\ge 3$ then
$2\cdot 3^n=2(1+2{)}^n=2[1+2n+\frac{n(n-1)}{2!}\cdot 2^2+o(n^2\left)\right]>2+4n^2=3n^2+(n^2+2)\ge 3n^2+11>3n^2+6$

$⟹{2.3}^n>3n^2+6$

Hence contradiction to (1)
Hence $n=1$ or $2$
If $n=1$, then $3^{2n}+3n^2+7=19$ and this is not a perfect square.
If $n=2$, we obtain $3^{2n}+3n^2+7=81+12+7=100={10}^2$
$\therefore n=2$ is the only solution.