## Task A-2.2.

Determine all pairs of natural numbers $(m, n)$ for which there exist integers $a, b$ and $c$ such that

$$
a+b+c=0 \quad \text{and} \quad a^{2}+b^{2}+c^{2}=2^{m} \cdot 3^{n}
$$
## Solution.

We arrive at the solution by analyzing the parity of the numbers $a, b$, and $c$.

Let $2^{k}$ be the highest power of 2 that divides all three numbers $a, b, c$, where $k \geqslant 0$. Write $a=2^{k} x, b=2^{k} y, c=2^{k} z$ for natural numbers $x, y, z$.

At least one of the numbers $x, y, z$ is odd. Given $x+y+z=0$, we conclude that exactly one of the numbers $x, y, z$ is even, and the other two are odd. Therefore, the expression $x^{2}+y^{2}+z^{2}$ gives a remainder of 2 when divided by 4, i.e., it is divisible by 2 but not by 4.

Since the second condition in the problem states $2^{2 k}\left(x^{2}+y^{2}+z^{2}\right)=2^{m} 3^{n}$, we conclude that $m=2 k+1$ for a non-negative integer $k$.

By constructing examples, we can see that all pairs where $m$ is odd and $n$ is any natural number are indeed achievable:

- $m=2 k+1, n=2 l$ is achieved with $a=2^{k} 3^{l}, b=-2^{k} 3^{l}, c=0$;
- $m=2 k+1, n=2 l+1$ is achieved with $a=b=2^{k} 3^{l}, c=-2^{k+1} 3^{l}$.