## Task A-4.3.

Determine all natural numbers $m$ and $n$ for which $2^{n}+5 \cdot 3^{m}$ is a square of some natural number.
## Second Solution.

As in the first solution, we conclude that $n=2 n_{1}, n_{1} \in \mathbb{N}$.

Now consider the remainders that the expressions in the initial equation give when divided by 8. If $n_{1} \geqslant 2$, the first addend on the left side is divisible by 8, and the second gives a remainder of 7 or 5, depending on the parity of $m$. The perfect square on the right side gives a remainder of 0 or 1, which leads to a contradiction. Therefore, the assumption $n_{1} \geqslant 2$ is incorrect, so $n_{1}=1$, or $n=2$.

The initial equality can be factored as:

$$
5 \cdot 3^{m}=(x-2)(x+2)
$$

As in the first solution, we conclude that the factors on the right side are relatively prime.

The smaller factor $x-2$ can be equal to $1, 3^{m}$, or 5.

In the first case, $x-2=1$ and $x+2=5 \cdot 3^{m}$. Directly, we see that $x=3$, and thus $x+2=5$, but this is not of the form $5 \cdot 3^{m}$ for any natural number $m$.

In the second case, $x-2=3^{m}$ and $x+2=5$. Again, $x=3$, and again we do not get a solution, since $x-2=1$ is not of the form $3^{m}$ for any natural number $m$.

In the last case, $x-2=5$ and $x+2=3^{m}$. From the first equation, $x=7$, and from the second, $9=3^{m}$, from which we get $m=2$.

Therefore, the only such pair of numbers $(m, n)$ is $(2,2)$.

Note: The parity of the number $n$ at the beginning of both solutions can be proven by considering the remainders when divided by 5.

As the scoring schemes suggest, finding the solution $(m, n)=(2,2)$ alone is worth 0 points.