Find all integers $x, y \geq 1$ such that $2^{x}-5=11^{y}$.

## Third Part

The following exercises, a bit more exotic, were not covered during the session.
First, $(x, y)=(4,1)$ is a solution. We assume that $y \geq 2, x \geq 4$. We apply the method of Dan Schwarz by rewriting the equation

$$
2^{4}\left(2^{x-4}-1\right)=11\left(11^{y-1}-1\right)
$$

Thus, $2^{4} \mid 11^{y-1}-1$. The order of 11 modulo $2^{4}$ is 4. Therefore, $61\left|11^{4}-1\right| 11^{y-1}-1$, so $61 \mid 2^{x-4}-1$. But the multiplicative order of 11 modulo $2^{4}$ is 4. Therefore, $61\left|11^{4}-1\right| 11^{y-1}-1$. The order of 2 modulo 61 is 60, so $41\left|2^{60}-1\right| 2^{x-4}-1$, which means $41 \mid 11^{y-1}-1$. The multiplicative order of 11 modulo 41 is 40, so $2^{5}\left|11^{40}-1\right| 11^{y-1}-1$, which means $2^{5} \mid 2^{4}\left(2^{x-4}-1\right)$. This is absurd and shows that there are no other solutions.