2.12. (GDR, 77). How many pairs of values $p, q \in \mathbf{N}$, not exceeding 100, exist for which the equation

$$
x^{5}+p x+q=0
$$

has solutions in rational numbers?
2.12. Let for some values

$$
p, q \in\{1 ; 2 ; \ldots ; 100\}
$$

the number $x \in \mathbf{Q}$ satisfies the equation

$$
x^{5}+p x+q=0
$$

Since all coefficients of the polynomial on the left side of the equation are integers, and the coefficient of the leading term is 1, by Theorem 60, any rational root of this polynomial, including the number $x$, is an integer. We will prove that $-30 < x < 3$.

If $x \geq 3$, then

$$
x^{5}+p x+q \geq 3^{5} + 3p + q > 0
$$

If $x \leq -3$, then

$$
x^{5}+p x+q \leq -3^{5} - 3p + q < 0
$$

Thus, the only possible integer values for $x$ are $-2, -1, 0, 1, 2$. For each of these values, we can find pairs $(p, q)$ such that the equation holds. If $x > 0$, then $q = -x^5 - px$, and since $p$ and $q$ are positive integers, $p$ must be less than or equal to 99. If $x < 0$, then $q = -x^5 - px$, and since $p$ and $q$ are positive integers, $p$ must be less than or equal to 34. If all the pairs obtained are distinct, the number of such pairs is $99 + 34 = 133$.