3. In the set of integers, solve the equation

$$
x^{5}+2=3 \cdot 101^{y}
$$
Solution. Let $(x, y)$ be a solution to the equation. If there exists $y_0$ such that $y=-k$, then the equation reads

$$
x^{5}+2=3 \cdot 101^{-k}
$$

If we multiply both sides by $100^{k}$, we get

$$
101^{k}\left(x^{5}+2\right)=3
$$

Since $101^{k}$ and $x^{2}+3$ are integers, and $k>0$, it follows from the last equation that $101|3$, which is a contradiction. Therefore, $y$ is a non-negative integer.

We will prove that $y=0$. Suppose $y \geq 1$. Then

$$
x^{5}+2=3 \cdot 101^{y} \equiv 0(\bmod 101)
$$

i.e., $x^{5} \equiv-2(\bmod 101)$, from which we get

$$
x^{100} \equiv\left(x^{5}\right)^{20} \equiv(-2)^{20}(\bmod 101)
$$

From the initial equation, it is clear that $x$ is not divisible by 101, so by Fermat's Little Theorem applied to the prime number 101, it follows that

$$
x^{100} \equiv 1(\bmod 101)
$$

Now, from $(-2)^{20}=\left(2^{10}\right)^{2}=1024^{2} \equiv 14^{2}(\bmod 101)$ and the above congruences, we have

$$
1 \equiv x^{100} \equiv 14^{2} \equiv 95(\bmod 101)
$$

which is a contradiction. From the obtained contradiction, it follows that we only need to consider the case $y=0$. Then the equation becomes $x^{5}+2=3$, from which we get $x^{5}=1$, i.e., $x=1$.

Finally, the pair $(x, y)=(1,0)$ is the only solution to the given equation.