# Positive integers $n$ that divide $\sigma_2(n)$

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 $$pq^2 \mid 1+p^2+q^2+q^4.$$

• If you look more closely, you need q^2 divides 1 + p^2 and p divides (1 + q^2 + q^4). This means there is k < p, k=3 mod 4, with kp = (1 + q^2 + q^4). I bet it is not far from this to a proof of impossibility, or a small counterexample. What have you tried? Gerhard "Ask Me About System Design" Paseman, 2011.01.11 – Gerhard Paseman Jan 11 '11 at 23:13
• Just a note that even without the mod 4 condition on p, there is only one example of such an n with p and q both < 7919, which suggests to me that the condition may not be needed to prove impossibility (for sufficiently large n). – Matthew Conroy Jan 12 '11 at 0:09
• sorry, k < (p^3)/4. Makes it more of a challenge, but still is limiting. Gerhard "Ask Me About System Design" Paseman, 2011.01.11 – Gerhard Paseman Jan 12 '11 at 3:09

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.

• alpoge--you need 50 rep shekels to comment on questions/answers not your own. See the faq under reputation. – drbobmeister Jan 12 '11 at 2:07
• This is a comment to Alex's post, since I can't actually comment on it (I complain a lot!): You mean $q\equiv 1\pmod{4}$, since $1\pmod{4}$'s are the ones that split in the Gaussian integers. – alpoge Jan 12 '11 at 4:04
• Of course, thanks! I must have slept upside down tonight. – Alex B. Jan 12 '11 at 4:09

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.