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 $$ pq^2 \mid 1+p^2+q^2+q^4. $$
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 $p$, $q$...
To state the obvious: Such an $n$ must naturally have $p\equiv 1 \pmod{12}$ and $q\equiv 1\pmod{4}$. The first condition is implied by the fact that $1 + q^2 + q^4 = 0\pmod{p}$, and so, since $p\neq 3$, $q^6\equiv 1\pmod{p}$ giving that $3\vert p-1$, or $p\equiv 1\pmod{3}$. The second is given by the fact that $1+p^2\equiv 0\pmod{q}$ gives that $p^2\equiv -1\pmod{q}$, and so $q\equiv 1\pmod{4}$, since $\left(\frac{-1}{q}\right) = 1$.
EDIT: Typos galore!
EDIT 2: I undeleted.
Here are some partial results. I hope that they will help narrow down the search for counterexamples. Suppose that $pq^2=n|\sigma_2(n)=1+p^2+q^2+p^2q^2+q^4$.
Claim 1: We must have $q\equiv 1\pmod 4$.
Proof: Suppose $q\equiv 3\pmod 4$. Then $q$ remains prime/irreducible in $\mathbb{Z}[i]$. So $q| (1 + p^2) = (1+ip)(1-ip)\Rightarrow q$ divides one of $1\pm ip$ in $\mathbb{Z}[i]$, which is impossible.
Claim 2: We must have $p\equiv1\pmod 3$.
Proof: We have $p|(1+q^2+q^4) = (1+q^2)^2 - q^2 = (1+q+q^2)(1-q+q^2)$, so $p$ divides one of these factors. Suppose that $p$ divides $1+q+q^2$. Then $\mathbb{F}_p$ contains the cube roots of unity, so $p\equiv1\pmod 3$. Similarly, if $p$ divides the second factor, then $\mathbb{F}_p$ contains the sixth roots of unity.
Note that for each such $p$, the last observation implies that there are at most two choices for $q$, for which the required divisibility can possibly hold. Namely, if $p\equiv1\pmod 3$, then there exist integers $A$ and $B$ such that $p=A^2 + 27B^2$ (this is due to Gauss) and the non-trivial cube roots of 1 mod $p$ are $$ \frac{A+9B}{A-9B}\text{ and }\frac{A-9B}{A+9B}. $$ Since $q$ is a cube root of 1 mod $p$ and since we also know that $q^2|(1+p^2)$, we have that $q$ is less than $p$ and so it must be one of the two numbers between 3 and $p-2$ satisfying the above congruences. Of course, often, these numbers won't be congruent to 1 mod 4 or won't be prime. Whether that's never the case, I am not sure.
