\section*{Problem 4 - 311024}

Determine all natural numbers \(t\) for which \(\sqrt{t+24 \sqrt{t}}\) is rational!
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Let \(r^{2}=t+24 \sqrt{t}\) with a rational number \(r\). Then \(\sqrt{t}=\frac{r^{2}-t}{24}\) is also a rational number, since \(t\) is a natural number. The square root of a natural number \(t\) is rational if and only if \(t\) is a perfect square, i.e., there exists a natural number \(n\) such that \(t=n^{2}\).

Then \(r^{2}=n^{2}+24 n\) is a natural number, and thus \(r\) is also a natural number. Therefore, \(r^{2}+144=n^{2}+2 \cdot n \cdot 12 + 12^{2}=(n+12)^{2}\), so with \(m:=n+12>0\), we have \(144=m^{2}-r^{2}=(m+r)(m-r)\). Since \(m>0\) and \(r>0\), \(m+r>0\) and thus \(m-r>0\), \(m+r\) and \(m-r\) are positive divisors and their respective counterparts of 144.

Since \(r>0\), \(m+r\) is the larger and \(m-r\) is the smaller divisor, and since \((m+r)-(m-r)=2 r\), the two divisors differ by a multiple of 2. Since 144 is divisible by 2, both factors must be divisible by 2. Since 144 has exactly the numbers 2, 4, 6, and 8 as even divisors that are less than \(\sqrt{144}=12\), the following solutions arise:
- For \(m-r=2\), \(m+r=72\), so \(r=35, m=37, n=m-12=25\), and \(t=n^{2}=625\). Indeed, \(\sqrt{625+24 \cdot \sqrt{625}}=\sqrt{625+24 \cdot 25}=\sqrt{1225}=35\) is rational.
- For \(m-r=4\), \(m+r=36\), so \(r=16, m=20, n=m-12=8\), and \(t=n^{2}=64\). Indeed, \(\sqrt{64+24 \cdot \sqrt{64}}=\sqrt{64+24 \cdot 8}=\sqrt{256}=16\) is rational.
- For \(m-r=6\), \(m+r=24\), so \(r=9, m=15, n=m-12=3\), and \(t=n^{2}=9\). Indeed, \(\sqrt{9+24 \sqrt{9}}=\sqrt{9+24 \cdot 3}=\sqrt{81}=9\) is rational.
- And for \(m-r=8\), \(m+r=18\), so \(r=5, m=13\), and \(n=t=1\). Indeed, \(\sqrt{1+24 \cdot \sqrt{1}}=\sqrt{25}=5\) is rational.

Thus, there are exactly four solutions \(t \in\{1,9,64,625\}\).