For a positive integer $n$ let 
$$
\sigma\_2(n) = \sum\_{d \mid n} d^2.
$$


There are many positive integers $n$ for which


$$
n \mid \sigma\_2(n).
$$


But, when $n$ has the particular form


$$
n=pq^2
$$


where $p, q$ are distinct odd prime numbers and
$$
p \equiv 1 \pmod{4} 
$$


there seems to be none.


Question: What happens in these case.


It is easy to see that the condition is equivalent to
$$
pq^2 \mid 1+p^2+q^2+q^4.
$$



Positive integers $n$ that divide $\sigma_2(n)$
As already mentioned in other posts, $p\equiv1\bmod3$ and $q\equiv1\pmod4$.
Since $p\mid q^4+q^2+1=(q^2+q+1)(q^2-q+1)$, the prime $p$ divides (at least)
one of the two factors; in particular, $p\le 1+q+q^2<(1+q)^2$ implying
$q>\sqrt p-1$. It happens very rare that $p^2+1$ is divisible by a square
$>(\sqrt p-1)^2$. Excluding the case $p=7$ and $q=5$ (when we indeed have
$q^2\mid p^2+1$ and $p\mid q^4+q^2+1$), there are only 19 such cases
for $p<10^8$; in all them the corresponding factor $q^2>(\sqrt p-1)^2$
involves $q$ *prime* such that $p\nmid q^4+q^2+1$. Even more,
the residue $q^4+q^2+1\pmod p$ seems to be a completely random number...


It looks like the two conditions $p^2\equiv-1\bmod{q^2}$ and $q^4+q^2+1\equiv0\bmod p$
do not "feel each other". Therefore, it is quite unlikely that a solution
to the system of congruences in prime $p>7$ and integer $q>5$ (not necessarily prime!)
exists. I even doubt about integer solutions $p$, $q$...


