Find all perfect squares $n$ such that if the positive integer $a \geqslant 15$ is some divisor of $n$ then $a+15$ is a prime power.
We call a positive integer \(a\) "nice" if \(a+15\) is a prime power.

From the definition, the numbers \(n=1,4,9\) satisfy the required property. Suppose that for some \(t \in \mathbb{Z}^{+}\), the number \(n=t^{2} \geqslant 15\) also satisfies the required property. We have two cases:

1. If \(n\) is a power of 2, then \(n \in\{16,64\}\) since

\[
2^{4}+15=31, \quad 2^{5}+15=47, \quad \text{and} \quad 2^{6}+15=79
\]

are prime, and \(2^{7}+15=143=11 \cdot 13\) is not a prime power. (Thus \(2^{7}\) does not divide \(n\) and therefore no higher power of 2 satisfies the required property.)

2. Suppose \(n\) has some odd prime divisor \(p\). If \(p>3\) then \(p^{2} \mid n\) and \(p^{2}>15\) which imply that \(p^{2}\) must be a nice number. Hence

\[
p^{2}+15=q^{m}
\]

for some prime \(q\) and some \(m \in \mathbb{Z}^{+}\). Since \(p\) is odd, then \(p^{2}+15\) is even, thus we can conclude that \(q=2\). I.e.

\[
p^{2}+15=2^{m}
\]

Considering the above modulo 3, we can see that \(p^{2}+15 \equiv 0,1 \bmod 3\), so \(2^{m} \equiv 1 \bmod 3\), and so \(m\) is even. Suppose \(m=2k\) for some \(k \in \mathbb{Z}^{+}\). So we have \((2^{k}-p)(2^{k}+p)=15\) and \((2^{k}+p)-(2^{k}-p)=2p \geqslant 10\). Thus

\[
2^{k}-p=1 \quad \text{and} \quad 2^{k}+p=15
\]

giving \(p=7\) and \(k=3\). Thus we can write \(n=4^{x} \cdot 9^{y} \cdot 49^{z}\) for some non-negative integers \(x, y, z\).

Note that 27 is not nice, so \(27 \nmid n\) and therefore \(y \leqslant 1\). The numbers 18 and 21 are also not nice, so similarly, \(x, y\) and \(y, z\) cannot both be positive. Hence, we just need to consider \(n=4^{x} \cdot 49^{z}\) with \(z \geqslant 1\).

Note that \(7^{3}\) is not nice, so \(z=1\). By checking directly, we can see that \(7^{2}+15=2^{6}, 2 \cdot 7^{2}+15=113, 4 \cdot 7^{2}+15=211\) are nice, but \(8 \cdot 7^{2}\) is not nice, so only \(n=49,196\) satisfy the required property.

Therefore, the numbers \(n\) which satisfy the required property are \(1,4,9,16,49,64\) and 196.

Remark by PSC. One can get rid of the case \(3 \mid n\) by noting that in that case, we have \(9 \mid n\). But then \(n^{2}+15\) is a multiple of 3 but not a multiple of 9 which is impossible. This simplifies a little bit the second case.