3. Find all pairs of positive integers $a$ and $b$, for which the number $a^{2}+b$ is 62 greater than the number $b^{2}+a$.
3. We will write the equation and immediately simplify the right side to a product:

$$
62=\left(a^{2}+b\right)-\left(b^{2}+a\right)=\left(a^{2}-b^{2}\right)-(a-b)=(a-b)(a+b-1) .
$$

The product of integers $u=a-b$ and $v=a+b-1$ is thus equal to the product of two prime numbers $2 \cdot 31$. Since $v \geq 1+1-1=1$, the number $u$ must also be positive and clearly $u<v$, so $(u, v)$ is one of the pairs $(1,62)$ or $(2,31)$. If we express $a, b$ in terms of $u, v$, we get

$$
a=\frac{u+v+1}{2} \quad \text{and} \quad b=\frac{v-u+1}{2} .
$$

For $(u, v)=(1,62)$, we obtain the solution $(a, b)=(32,31)$, and the pair $(u, v)=(2,31)$ corresponds to the second solution $(a, b)=(17,15)$. The problem has no other solutions.

For a complete solution, award 6 points. For guessing both solutions, award 1 point (i.e., for guessing only one solution, award no points). For providing the factorization, award 2 points, and if the solver is aware that it suffices to examine only a finite number of possibilities, add another 1 point.