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


Solution: If $3^{2 n}+3 n^{2}+7=b^{2}$ for some natural number $b$, then $b^{2}>3^{2 n}$ so that $b>3^{n}$. This implies that $b \geq 3^{n}+1$. Thus

$$
3^{2 n}+3 n^{2}+7=b^{2} \geq\left(3^{n}+1\right)^{2}=3^{2 n}+2 \cdot 3^{n}+1
$$

This shows that $2 \cdot 3^{n} \leq 3 n^{2}+6$. If $n \geq 3$, this cannot hold. One can prove this eithe by induction or by direct argument:

If $n \geq 3$, then

$$
\begin{aligned}
2 \cdot 3^{n}=2(1+2)^{n}=2\left(1+2 n+(n(n-1) / 2) \cdot 2^{2}\right. & +\cdots)>2+4 n+4 n^{2}-4 n \\
& =3 n^{2}+\left(n^{2}+2\right) \geq 3 n^{2}+11>3 n^{2}+6
\end{aligned}
$$

Hence $n=1$ or 2 .

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

