## Task A-3.2.

Determine all pairs of integers $(m, n)$ such that $4 \cdot 3^{2 m}+5=n^{2}$.
## First solution.

Notice that for $m<0$ the number $4 \cdot 3^{2 m}$ is not an integer, while $n^{2}-5$ is, so in this case the equation has no solution.

(2 points)

For $m=0$ we have $n^{2}=4 \cdot 1+5=9$, from which it follows that $n=3$ or $n=-3$.

In this case, we have two solutions: $(0,3)$ and $(0,-3)$.

For $m \geqslant 1$, the number $4 \cdot 3^{2 m}$ is divisible by 3,

so the number $n^{2}=4 \cdot 3^{2 m}+5$ gives a remainder of 2 when divided by 3,

but this is not possible, since the square of an integer when divided by 3 can only give remainders of 0 or 1. Therefore, for $m \geqslant 1$ the equation has no solutions.

(3 points)