Solve the equation $3^{x}+4^{y}=5^{z}$ in the set of positive integers.
Solution. Since $5^{z}$ gives a remainder of 1 when divided by 4, $3^{x}$ must also have a remainder of 1, which means $x$ is even: $x=2 X$. The remainder of $4^{y}$ when divided by 3 is 1, so the remainder of $5^{z}$ when divided by 3 is also 1, hence $z$ is even: $z=2 Z$. Accordingly, rearranging and factoring the equation:

$$
2^{2 y}=4^{y}=5^{z}-3^{x}=5^{2 Z}-3^{2 X}=\left(5^{Z}+3^{X}\right)\left(5^{Z}-3^{X}\right)
$$

Since the value of the product is $2^{2 y}$, a power of 2, both $5^{Z}+3^{X}$ and $5^{Z}-3^{X}$ must also be powers of 2: for suitable integers $a>b$, we have $5^{Z}+3^{X}=2^{a}$ and $5^{Z}-3^{X}=2^{b}$. Solving for $5^{Z}$ and $3^{X}$ from the two equations:

$$
5^{Z}=\frac{2^{a}+2^{b}}{2}=2^{a-1}+2^{b-1}, \quad 3^{X}=2^{a-1}-2^{b-1}
$$

Since $5^{Z}$ and $3^{X}$ are integers, $a$ and $b$ must be positive integers. If $2^{a-1}-2^{b-1}=2^{b-1}\left(2^{a-b}-1\right)$ is a power of 3, then $2^{b-1}$ must also be a power of 3, which is only possible if $b=1$. Therefore, $3^{X}=2^{a-1}-1$, so the remainder of $2^{a-1}$ when divided by 3 is 1, which means $a-1$ is even: $a-1=2 A$. Factoring the right-hand side of the previous equation:

$$
3^{X}=2^{2 A}-1=\left(2^{A}+1\right)\left(2^{A}-1\right)
$$

Thus, $2^{A}+1$ and $2^{A}-1$ are powers of 3 whose difference is 2, so $2^{A}+1=3$ and $2^{A}-1=1$. From the above equations, it follows that

$$
A=1, \quad a=3, \quad 3^{X}=3, \quad 5^{Z}=5, \quad 4^{y}=(5+3)(5-3)=16
$$

From which we get

$$
x=2 X=2, \quad y=2, \quad z=2 Z=2
$$

Since $3^{2}+4^{2}=5^{2}$ is indeed true, the only solution to the equation is $x=y=z=2$.