## Task 16/81

Determine all ordered pairs of natural numbers $(n ; m)$ that satisfy the equation $2^{n}+65=m^{2}$!
For odd $n$, $2^{n} \equiv \pm 2(\bmod 10)$, so $2^{m}+65 \equiv (5 \pm 2)(\bmod 10)$ and thus is not a square of a natural number [for the squares of natural numbers, it is well-known that one of the congruences $\left.m^{2} \equiv 0 ; 1 ; 4 ; 5 ; 6 ; 9(\bmod 10)\right]$ always holds]. Therefore, there is no solution for odd $n$.

Now let $n=2 k$ with $k \in \mathbb{N}$. Then

$$
2^{n}5$ follows

$$
\left(2^{k}\right)^{2}+2 \cdot 2^{5}+15$ the following chain of inequalities:

$$
2^{n}=\left(2^{k}\right)^{2}<2^{n}+65<\left(2^{k}+1\right)^{2}
$$

Thus, $m^{2}=2^{n}+65$ lies between the squares of two consecutive natural numbers and therefore cannot itself be the square of a natural number.

The examination of the cases $k=0 ; 1 ; 2 ; 3 ; 4 ; 5$ yields the two solutions $\left(n_{1}=2 k_{1} ; m_{1}\right)=(4 ; 9)$, $\left(n_{2}=2 k_{2} ; m_{2}\right)=(10 ; 33)$. The verification confirms the correctness.