\section*{Problem 5 - 161245}

Determine the number of all pairs \((p, q)\) of natural numbers with \(1 \leq p \leq 100\) and \(1 \leq q \leq 100\) and the property that the equation \(x^{5} + p x + q = 0\) has at least one rational solution.
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Assume for a pair \((p, q)\) of natural numbers with \(1 \leq p \leq 100\) and \(1 \leq q \leq 100\) that \(x\) is a rational solution of the equation \(x^{5} + p x + q = 0\).

Since \(p \geq 1, q \geq 1\), it follows that \(x = -\frac{a}{b} < 0\), where \(a, b\) are coprime numbers with \(a \neq 0, b \neq 0\). Therefore,

\[
-\frac{a^{5}}{b^{5}} - p \frac{a}{b} + q = 0 \quad \Rightarrow \quad q b^{5} = p a b^{4} + a^{5}
\]

Hence, \(b \mid a^{5}\), so \(b = 1\) because \(a\) and \(b\) are coprime. This yields:

\[
q = p a + a^{5} \quad \Rightarrow \quad q = a(p + a^{4})
\]

Since \(p, q, a \geq 1\) and \(q \leq 100\), it follows that \(1 \leq a < 3\), so \(a = 1\) or \(a = 2\).

1. In the case \(a = 1\), we get \(q = p + 1\).

Since \(q \leq 100\), it follows that \(p \leq 99\), meaning there are 99 pairs \((p, p+1) (p = 1, 2, \ldots, 99)\) that could have the required property.

2. In the case \(a = 2\), we get \(q = 2p + 32\).

Since \(q \leq 100\), it follows that \(p \leq 34\), meaning there are 34 pairs \((p, 2p + 32) (p = 1, 2, \ldots, 34)\) that could have the required property.

The 99 + 34 = 133 pairs are all distinct because \(p + 1 < 2p + 32\). Furthermore, they have the required property; indeed,

in case 1: \(x = -\frac{a}{b} = -1\), so \(x^{5} + p x + (p + 1) = 0\) and

in case 2: \(x = -\frac{a}{b} = -2\), so \(x^{5} + p x + (2p + 32) = 0\).

Therefore, there are exactly 133 pairs \((p, q)\) with the required property.

\section*{Adapted from \([5]\)}