Are there any integer solutions to $\gcd(\sigma(n), \sigma(n^2)) = 1$ other than for prime $n$? 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!
 A: 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 $\sigma(p^k)$. 
A: 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
$$gcd(p+1,q^2+q+1)=\gcd(q+1,p^2+p+1)=1 \,.$$
A: 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 $n$.
