Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Here is my question:
Are there infinitely many primes $q$, such that $\sigma(p^k) = 2q$, where $p$ is prime and $p \equiv k \equiv 1 \pmod 4$?
MY ATTEMPT
Since $p$ is prime, and $k \geq 1$ is an integer (satisfying $k \equiv 1 \pmod 4$), then we get $$q=\frac{\sigma(p^k)}{2}=\frac{p^{k+1}-1}{2(p-1)}=\Bigg(\frac{p^{(k+1)/2} + 1}{2}\Bigg)\cdot\Bigg(\frac{p^{(k+1)/2} - 1}{p - 1}\Bigg)=\Bigg(\frac{p^{(k+1)/2} + 1}{2}\Bigg)\cdot\sigma(p^{(k-1)/2}),$$ whereupon we obtain, since $q$ is prime, that either $$\text{Case (1): } \frac{p^{(k+1)/2} + 1}{2} = 1$$ XOR $$\text{Case (2): } \sigma(p^{(k-1)/2}) = 1$$ holds.
We then get, from the first condition, that $$\sigma(p^{(k-1)/2}) = q$$ which, together with $\sigma(p^k)=2q$, implies that $$\sigma(p^k) = 2\sigma(p^{(k-1)/2})$$ Dividing both sides by $p^k$, and noting that $p^k = (p^{(k-1)/2})(p^{(k+1)/2})$, we obtain $$I(p^k) = \Bigg(\frac{2}{p^{(k+1)/2}}\Bigg)I(p^{(k-1)/2})$$ But since $p \equiv k \equiv 1 \pmod 4$, then $$1 < I(p^k) < \frac{5}{4}$$ and $$1 \leq I(p^{(k-1)/2}) < \frac{5}{4}.$$
This implies that $$\frac{4}{5} < \frac{2}{p^{(k+1)/2}} = \frac{I(p^k)}{I(p^{(k-1)/2})} < \frac{5}{4}$$ which means that $$\frac{2}{5} < \frac{1}{p^{(k+1)/2}} < \frac{5}{8},$$ further giving $$\frac{8}{5} < p^{(k+1)/2} < \frac{5}{2},$$ which implies that $$\frac{64}{25} < p^{k+1} < \frac{25}{4}.$$
We get $$\frac{39}{25} < p^{k+1} - 1 < \frac{21}{4}.$$ But since $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$, then $$p - 1 \geq 4.$$ In particular, we obtain the upper bound $$q = \frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} < \frac{21}{32},$$ which contradicts the requirement that $q$ must be prime.
We then get, from the second condition, that $$k = 1,$$ and therefore that $$\frac{p+1}{2}=q$$ where both $p$ and $q$ are primes.
Alas, this is where I get stuck!
It seems that it is not known if there are infinitely many primes $q$ such that $2q−1$ is also prime according to OEIS/A005382 and Solve $2p=q+1$ where $p$ and $q$ are prime.