Does this equation have any solutions for prime $q$ and $k > 1$, $k \equiv 1 \pmod 4$?

Question

Does this equation have any solutions for prime $q$ and $k > 1$, $k \equiv 1 \pmod 4$?

$$\sigma\left(\frac{1}{2}\sigma(q^k)\right) = 2q$$

Here $\sigma$ is the classical sum-of-divisors function. So for example, $$\sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28.$$

Answer 1

There are no solutions for $k > 1$ and $k \equiv 1 \pmod{4}$. We have $\sigma(n) \geqslant n$, where the inequality is strict for $n > 1$, hence

$$\sigma\biggl(\frac{1}{2}\sigma(q^k)\biggr) \geqslant \frac{1}{2}\sigma(q^k) \geqslant \frac{1}{2}q^k,$$

so we would need

$$\frac{1}{2}q^{k-1} \leqslant 2.$$

But for $k \equiv 1 \pmod{4}$, we have $q^{k-1} \geqslant q^4 \geqslant 16$.

Answer 2

The restriction $k > 1$ is not needed. Consider the case $k=1$. If $q$ is prime, $\sigma(q) = q+1$ so you want $$ \sigma((q+1)/2) = 2 q $$ Note that $q$ must be odd for $(q+1)/2$ to be an integer.

Now if $(q+1)/2$ had two or more distinct prime factors, since $\sigma$ is multiplicative $\sigma((q+1)/2)$ would be the product of two integers $> 2$, which is not true for $2 q$. So $(q+1)/2$ must be a power of a single prime, say $p^m$. But $\sigma(p^m) = (p^{m+1}-1)/(p-1)$ and $2 q = 2 p^m - 1$, so we'd have $$ (2 p^m - 1)(p-1) = p^{m+1}-1 $$ which simplifies to $$ (p-2)(p^m-1) = 0$$ Thus either $m=0$ (which is impossible) or $p=2$ (also impossible, because it makes $2q = 2 p^m-1$ odd).

For solutions with $k=1$ where $q$ is not required to be prime, see OEIS sequence A113791.