4. Solve the equation $2^{x}+5^{y}+19=30^{z}$ in natural numbers.
Answer: $(8,4,2)$.

Solution. The right-hand side of the equation is divisible by 5, and the left-hand side gives the same remainder when divided by 5 as $2^{x}-1$. Therefore, $x$ is a multiple of 4, and in particular, $x \geqslant 4$. Moreover, the right-hand side is divisible by 3, and the left-hand side gives the same remainder when divided by 3 as $(-1)^{x}+(-1)^{y}+1=(-1)^{y}+2$. Therefore, $y$ is even, that is, $y=2 n$. Notice that

$$
2^{x}+5^{y}+19=2^{x}+5^{2 n}-1+20=2^{x}+\left(5^{n}-1\right)\left(5^{n}+1\right)+20
$$

The number $\left(5^{n}-1\right)\left(5^{n}+1\right)$ is divisible by 8 as the product of two consecutive even numbers. Then the left-hand side of the equation is divisible by 4 but not by 8, and the same must be true for $30^{z}$. Therefore, $z=2$ and $2^{x}+25^{n}=881$. For $n$, only the values 1 and 2 are possible. Only $n=2$ works, which gives $y=4$ and $x=8$.