4. Solve the equation $2^{x}+3^{y}+3=10^{z}$ in natural numbers.
Answer: $(2,1,1),(4,4,2)$.

Solution. The left side of the equation, when divided by 3, gives the same remainder as $(-1)^{x}$, while the remainder of the right side for any $z$ is 1. Therefore, $x$ is even and, in particular, $x \geqslant 2$. If $z=1$, then the equation is satisfied by $x=2, y=1$ and only they. Let $z \geqslant 2$. Then the right side of the equation is divisible by 4, and the left side, when divided by 4, gives the same remainder as $(-1)^{y}+3$. Therefore, $y$ is even, i.e., $y=2n$. We consider two cases.

1) $z=2$. We get $2^{x}+9^{n}=97$. For $n$, only the values 1 and 2 are possible. Only $n=2$ works, which gives $y=4$ and $x=4$.
2) $z \geqslant 3$. Then the right side of the equation is divisible by 8, and the left side is

$$
2^{x}+3^{2n}-1+4=2^{x}+(3^{n}-1)(3^{n}+1)+4
$$

Note that $(3^{n}-1)(3^{n}+1)$ is divisible by 8 as the product of two consecutive even numbers. Therefore, $2^{x}$ cannot be divisible by 8, which implies $x=2$. Thus, $9^{n}+7=10^{z}$. However, this is impossible, since the remainder of the right side when divided by 9 is 1 for any $z$, while the remainder of the left side is 7.