Are there any integer solutions to $\gcd(\sigma(n), \sigma(n^2)) = 1$ other than for prime $n$?

Question

A good day to everyone!

Are there any integer solutions to $\gcd(\sigma(n), \sigma(n^2)) = 1$ other than for prime $n$ (where $\sigma = \sigma_1$ is the sum-of-divisors function)?

Note that, if $n = p$ for prime $p$ then

$$\sigma(p) = p + 1$$ $$\sigma(p^2) = p^2 + p + 1 = p(p + 1) + 1.$$

These two equations can be put together into one as

$$\sigma(p^2) = p\sigma(p) + 1,$$

from which it follows that

$$\sigma(p^2) + (-p)\cdot\sigma(p) = 1.$$

The last equation implies that $\gcd(\sigma(p), \sigma(p^2)) = 1$.

I now attempt to show that prime powers also satisfy the number-theoretic equation in this question.

If $n = q^k$ for $q$ prime, then

$$\sigma(q^{2k}) = \frac{q^{2k + 1} - 1}{q - 1} = \frac{q^{2k + 1} - q^{k + 1}}{q - 1} + \frac{q^{k + 1} - 1}{q - 1} = \frac{q^{k + 1}(q^k - 1)}{q - 1} + \sigma(q^k).$$

Re-writing the last equation, we get

$$(q - 1)\left(\sigma(q^{2k}) - \sigma(q^k)\right) = q^{k + 1}(q^k - 1).$$

Since $\gcd(q - 1, q) = 1$, then we have

$$q^{k + 1} \mid \left(\sigma(q^{2k}) - \sigma(q^k)\right).$$

But we also have

$$\sigma(q^{2k}) - \sigma(q^k) = q^{k + 1} + q^{k + 2} + \ldots + q^{2k} \equiv 0 \pmod {q^{k + 1}}.$$

Alas, this is where I get stuck. (I know of no method that can help me express $1$ as a linear combination of $\sigma(q^{2k})$ and $\sigma(q^k)$, from everything that I've written so far.)

Anybody else here have any ideas?

Thank you!

Answer 1

Yes, $\sigma(p^k)$ and $\sigma(p^{2k})$ are relatively prime if $p$ is prime. It is I think best not to sum the geometric series. We have $$\sigma(p^k)=1+p+p^2+\cdots+p^k,$$ and $$\sigma(p^{2k})=1+p+p^2+\cdots +p^k+p^{k+1}+\cdots +p^{2k}.$$ The second sum is $$\left(1+p+\cdots+p^k\right)+\left(p^k+p^{k+1}+\cdots+p^{2k}\right)-p^k$$ (we added and subtracted $p^k$). Thus $$\sigma(p^{2k})=(1+p^k)\sigma(p^k)-p^k.$$ So the gcd of $\sigma(p^k)$ and $\sigma(p^{2k})$ is the same as the gcd of $p^k$ and $1+p+\cdots+p^k$. This is is obviously $1$, since the only prime that divides $p^{k}$ is $p$, and $p$ does not divide $\sigma(p^k)$.

Answer 2

If $n=p^k$ then $\sigma(n)=\dfrac{p^{k+1}-1}{p-1}$ and $\sigma(n^2)=\dfrac{p^{2k+1}-1}{p-1}$. So we'd want $\gcd(p^{k+1}-1,p^{2k+1}-1)=p-1$.

But in general, $$\gcd(a^m-1,a^n-1)= a^{\gcd(m,n)}-1$$

So we have $$\gcd(p^{k+1}-1,p^{2k+1}-1)=p^{\gcd(k+1,2k+1)}-1 = p-1$$

So this is true for any prime powers $n$.

Answer 3

Let $n=pq$ where $p,q$ are two distinct primes.

Then

$$\sigma(n)=(p+1)(q+1)$$ $$\sigma(n^2)=(1+p+p^2)(1+q+q^2)$$

$$\sigma(6)=12$$ $$\sigma(6^2)=7 \cdot 13 \,.$$

Note that as long as $1+p+p^2$ and $1+q+q^2$ are primes, then for $n=pq$ we have $gcd(\sigma(n), \sigma(n^2))=1$.

Actually you only need

$$gcd(p+1,q^2+q+1)=\gcd(q+1,p^2+p+1)=1 \,.$$