3. Find all natural numbers $a$ and $b$ such that

$$
a^{3}-b^{3}=633 \cdot p
$$

where $p$ is some prime number.
Answer: $a=16, b=13$.

Solution: Factorize the left side of the equation

$$
\begin{gathered}
(a-b)\left(a^{2}+a b+b^{2}\right)=3 \cdot 211 \cdot p \Longrightarrow \\
\Longrightarrow(a-b)\left((a-b)^{2}+3 a b\right)=3 \cdot 211 \cdot p
\end{gathered}
$$

Notice that the right side is divisible by 3, so at least one of the factors is divisible by 3. If

$$
a-b \vdots 3 \Longrightarrow(a-b)^{2}+3 a b \vdots 3
$$

and conversely

$$
(a-b)^{2}+3 a b: 3 \Longrightarrow(a-b)^{2}: 3 \Longrightarrow a-b: 3
$$

We get that the left side of the equation is divisible by 9, so $3 \cdot 211 \cdot p: 9$. Therefore, $p: 3$, and since $p \in \mathbb{P}$, then $p=3$.

In the end, we get the following equation

$$
(a-b)\left((a-b)^{2}+3 a b\right)=3^{2} \cdot 211
$$

where both factors are positive and divisible by 3. Also, note that the second factor is greater than the first, so there is only one way to factorize

$$
\left\{\begin{array}{l}
a-b=3 \\
(a-b)^{2}+3 a b=3 \cdot 211
\end{array}\right.
$$

Express $a$ and substitute into the second equation

$$
9+3(b+3) b=3 \cdot 211 \Longrightarrow b^{2}+3 b-208=0
$$

Solve the quadratic equation and get the roots

$$
b_{1}=\frac{-3-29}{2}=-16 \quad \text { and } \quad b_{1}=\frac{-3+29}{2}=13
$$

We are only interested in natural numbers, so $a=16$ and $b=13$.