## Task A-4.4. (4 points)

Determine all pairs of natural numbers $(m, n)$ such that $m^{5} + n^{2} = 1700$.
## Solution.

Since $n^{2}=1700-m^{5}$ is a non-negative number, we have $m^{5} \leqslant 1700<3125=5^{5}$, which implies that $m<5$.

If $m=1$, we get $n^{2}=1699$, which is impossible $\left(41^{2}<1699<42^{2}\right)$.

If $m=2$, we get $n^{2}=1668$, which is impossible $\left(40^{2}<1668<41^{2}\right)$.

If $m=3$, we get $n^{2}=1457$, which is impossible $\left(38^{2}<1457<39^{2}\right)$.

Therefore, $m$ cannot be 1, 2, or 3.

If $m=4$, we get $n^{2}=676$ and thus $n=26$.

The only solution is $(m, n)=(4,26)$.

Note. It can be used that a perfect square never ends in the digits 7 or 8.