Let's determine those natural numbers $x$ and $y$ whose sum is greater than 40 and which satisfy the following equation:

$$
2^{10 x+24 y-493}+1=9 \cdot 2^{5 x+12 y-248}
$$
Our equation can also be written as

$$
2^{3} \cdot 2^{10 x+4 y-496}-9 \cdot 2^{5 x+12 y-248}+1=0
$$

Let $2^{5 x+12 y-248}=p$, then

$$
8 p^{2}-9 p+1=0
$$

from which

$$
p_{1}=1=2^{0}, \quad \text { and } \quad p_{2}=\frac{1}{8}=2^{-3}
$$

Therefore, either

$$
5 x+12 y-248=0
$$

or

$$
5 x+12 y-248=-3
$$

Determine the integer solution of the indeterminate equation (1).

$$
\begin{aligned}
& x=\frac{248-12 y}{5}=49-2 y+\frac{3-2 y}{5}=49-2 y+t \\
& y=\frac{3-5 t}{2}=1-2 t+\frac{1-t}{2}=1-2 t+u \\
& t=1-2 u
\end{aligned}
$$

Substituting back

$$
\begin{aligned}
& x=1-2+4 u+u=5 u-1 \\
& y=49-10 u+2+1-2 u=52-12 u
\end{aligned}
$$

Thus $x+y=51-7 u$, and according to the problem (considering 0 as a natural number)

$$
0 \leqq 5 u-1 \quad 0 \leqq 52-12 u \quad \text { and } \quad 4040
$$

From these three inequalities, respectively,

$$
u \leqq \frac{49}{12}, \quad u \geqq 0 \quad u<\frac{9}{7}
$$

These inequalities are satisfied by $u=0$, and $u=1$.

Thus, we have two more solutions:

$$
x_{2}=49, \quad y_{2}=0
$$

and

$$
x_{3}=47-19=37, \quad y_{3}=5
$$

István Kovács (Kecskemét, Ságvári specialized college)