5. How many pairs of positive integers $p, q$ not exceeding 100 make the equation $x^{5}+p x+q=0$ have a rational solution?
5. Since $p, q \in\{1,2, \cdots, 100\}, x \in \mathrm{Q}$, then $x \in Z$.

If $x \geqslant 0$, then $x^{5}+p x+q \geqslant 1$. If $x \leqslant-3$, then $x^{5}+p x+q \leqslant-3^{5}-3+100<$ 0 . Therefore, only $x=-1$, or $x=-2$.

When $x=-1$, $q=p+1$, in this case there are exactly 99 pairs $(p, q)$ that satisfy the condition; when $x=-2$, $p$ $=2 q+32$. In this case, $p \leqslant 34$, there are exactly 34 pairs $(p, q)$ that satisfy the condition.
In summary, there are 133 pairs $(p, q)$ that satisfy the problem.