4. Solve the equation $2^{x}+6^{y}+5=15^{z}$ in natural numbers.
Answer: $(2,1,1),(2,3,2)$.

Solution. The right side of the equation is divisible by 3, and the left side gives the same remainder when divided by 3 as $(-1)^{x}+2$. Therefore, $x$ is even and, in particular, $x \geqslant 2$. We can write the equation as $2^{x}+6^{y}=15^{z}-5$ and consider two cases.

1) $z$ is odd. The remainder of $15^{z}-5$ when divided by 4 is the same as that of $(-1)^{z}+3$, which is 2. Then $6^{y}$ is not divisible by 4, so $y=1$ and $2^{x}=15^{z}-11$. The remainder of $15^{z}-11$ when divided by 8 is the same as that of $(-1)^{z}+5$, which is 4. This is only possible when $x=2$. Since $2^{2}+6^{1}+5=15^{1}$, we have the solution $(2,1,1)$.
2) $z$ is even. The remainder of $15^{z}-5$ when divided by 16 is the same as that of $(-1)^{z}+11$, which is 12. Note that the remainders of $6^{y}$ when divided by 16 can only be $6,4,8$, and 0, and none of these is 12. Therefore, $2^{x}$ cannot be divisible by 16, so $x=2$. Then the remainder of $6^{y}$ when divided by 16 must be 4, which is only possible when $y=2$. Since $2^{2}+6^{3}+5=15^{2}$, we have the solution $(2,3,2)$.