\section*{Problem 2 - 171042}

Determine all rational numbers \(x\) for which the number \(z=x^{2}+x+6\) is the square of a natural number!
Assume a rational number \(x\) has the required property. Then there exist integers \(p, q\) that are coprime with \(q>0\) and \(x=\frac{p}{q}\), and a natural number \(n\) such that

\[
\frac{p^{2}}{q^{2}}+\frac{p}{q}+6=n^{2}
\]

This implies \(p^{2}=q\left(-p-6 q+n^{2} q\right)\). Therefore, \(p^{2}\) is divisible by \(q\). If \(q\) were divisible by a prime number, this prime would have to be a factor of \(p^{2}\) and thus of \(p\), contradicting the coprimality of \(p\) and \(q\). Hence, \(q=1\) and we have:

\[
\begin{aligned}
p^{2}+p+6 & =n^{2} \\
\left(p+\frac{1}{2}\right)^{2} & =n^{2}-\frac{23}{4} \\
23 & =4 n^{2}-(2 p+1)^{2}=(2 n-2 p-1)(2 n+2 p+1)
\end{aligned}
\]

Thus, the prime number 23 is factored into two integers whose sum is a non-negative number, namely \(4 n\). Consequently, the second of the two possible integer factorizations \(23=1 \cdot 23=(-1) \cdot (-23)\) is ruled out, and we have either

\[
2 n-2 p-1=1, \quad 2 n+2 p+1=23 \quad \text { or } \quad 2 n-2 p-1=23, \quad 2 n+2 p+1=1
\]

In the first case, \(n-p=1, n+p=11\) and thus \(p=5\), in the second case, \(n-p=12, n+p=0\) and thus \(p=-6\).

Therefore, only the numbers \(x=5\) and \(x=-6\) can have the required properties. Indeed, both \(25+5+6=36\) and \(36-6+6=36\) are squares of natural numbers.