(given to Jean-Alix David and William Fujiwara). Find all natural numbers $x, y$ and $z$ satisfying $3^{x}+4^{y}=5^{z}$.
. By looking modulo 3, we show that $z=2 z^{\prime}$ must be an even number. Similarly, by looking modulo 4, we show that $x=2 x^{\prime}$ must also be even. The equation then becomes:

$$
3^{2 x^{\prime}}=\left(5^{z^{\prime}}-2^{y}\right)\left(5^{z^{\prime}}+2^{y}\right)
$$

Each of the factors must be a power of 3, but their sum is not divisible by 3. One of the two, necessarily the smaller one (here the first), must be equal to 1. This gives us the two equations $5^{z^{\prime}}-2^{y}=1$ and $5^{z^{\prime}}+2^{y}=3^{2 x^{\prime}}$. The first equation implies, by looking modulo 3, that $z^{\prime}$ is odd and that $y$ is even. But if $y \geqslant 3$, either of the equations implies, by looking modulo 8, that $z^{\prime}$ is even. Therefore, there is no solution with $y \geqslant 3$. The only remaining possibility is $y=2$, which directly gives $x=z=2$ as well.