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This is related to this earlier MSE question. In particular, it appears that there is already a proof for the equivalence $$\sigma(q^{k-1}) \text{ is a square } \iff k = 1.$$

Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.

Here is my question:

If $q$ is prime, can $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ be both squares when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$?

MY ATTEMPT

Suppose that $$\sigma(q^{k-1}) = a^2$$ and $$\frac{\sigma(q^k)}{2} = b^2$$ for $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$.

Since $\sigma(q^k) = q^k + \sigma(q^{k-1})$, it follows that $$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) = q^k + a^2.$$

Additionally, congruence-wise we obtain $$a^2 = \sigma(q^{k-1}) \equiv 1 + (k-1) \equiv k \equiv 1 \pmod 4,$$ from which it follows that $a$ is odd, and $$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) \equiv 1^1 + 1 \equiv 2 \pmod 4,$$ which implies that $b$ is likewise odd.

Now, using the definition of $\sigma(q^k)$ and $\sigma(q^{k-1})$ for $q$ prime, we derive $$\frac{1}{2}\cdot\frac{q^{k+1} - 1}{q - 1} = b^2$$ and $$\frac{q^k - 1}{q - 1} = a^2.$$

Assume to the contrary that $$\frac{1}{2}\cdot\frac{q^{k+1} - 1}{q - 1} = b^2 \leq a^2 = \frac{q^k - 1}{q - 1}.$$ This assumption leads to $$q^{k+1} - 1 \leq 2(q^k - 1)$$ which implies that $$16 = {5^1}(5-2) + 1 \leq q^k(q - 2) + 1 = q^{k+1} - 2q^k + 1 \leq 0,$$ since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$. This results in the contradiction $16 \leq 0$. Consequently, we conclude that $a < b$.

Furthermore, I know that $$(q+1) = \sigma(q) \mid \sigma(q^k) = 2b^2$$ so that $$\frac{q+1}{2} \leq b^2.$$

Finally, I also have $$\frac{q^{k+1} - 1}{2b^2} = q - 1 = \frac{q^k - 1}{a^2}.$$

Alas, here is where I get stuck.

CONJECTURE (Open)

If $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$, then $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ are both squares when $k = 1$.

SUMMARY OF RESULTS SO FAR

zongxiangyi appears to have proven the implication $$\sigma(q^k)/2 \text{ is a square} \implies k = 1.$$

The proof of the following implication is trivial $$k = 1 \implies \sigma(q^{k-1}) \text{ is a square}.$$ The truth value of the following implication is currently unknown: $$\sigma(q^{k-1}) \text{ is a square} \implies k = 1.$$

Together, the two results give $$\sigma(q^k)/2 \text{ is a square} \implies k = 1 \iff \sigma(q^{k-1}) \text{ is a square},$$ so that $\sigma(q^{k-1})$ is a square if $\sigma(q^k)/2$ is a square.

Therefore, $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ are both squares (given $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$) when $\sigma(q^k)/2$ is a square.

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  • $\begingroup$ I guess, If the conjecture holds, then $q\equiv 1 \pmod{2^n}$ for $n\ge 1$ such that $q>2^n$ . $\endgroup$ – Zongxiang Yi Jul 6 at 7:19
  • $\begingroup$ From the equation $$\sigma(q^k)/2 = b^2$$ I can derive $$\bigg(\frac{q^{(k+1)/2} + 1}{2}\bigg)\cdot\bigg(\frac{q^{(k+1)/2} - 1}{q - 1}\bigg) = b^2.$$ Note that $$\gcd\bigg(\frac{q^{(k+1)/2} + 1}{2},\frac{q^{(k+1)/2} - 1}{q - 1}\bigg)=1.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 7 at 3:35
  • $\begingroup$ Therefore, if $\sigma(q^k)/2$ is a square, then both $$\frac{q^{(k+1)/2} + 1}{2}$$ and $$\frac{q^{(k+1)/2} - 1}{q - 1}$$ are squares. In particular, $$\sigma(q^{(k-1)/2}) = \frac{q^{(k+1)/2} - 1}{q - 1}$$ is a square. So we now have the (somewhat stringent) requirement that $$\sigma(q^k)/2, \sigma(q^{k-1}), \text{ and } \sigma(q^{(k-1)/2})$$ are all squares. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 7 at 3:45
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Here are a couple of other approaches to consider which may be useful. First, your equation of

$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) = q^k + a^2 \tag{1}\label{eq1}$$

can be rewritten as

$$2b^2 - a^2 = q^k \tag{2}\label{eq2}$$

This is in the generalized Pell equation form of $x^2 - Dy^2 = N$. The blog Solving the generalized Pell equation explains how to solve this.

Next, note that

$$\sigma(q^{k-1}) = \sum_{i=0}^{k-1} q^i \tag{3}\label{eq3}$$

$$\sigma(q^{k}) = \sum_{i=0}^{k} q^i \tag{4}\label{eq4}$$

Thus, you can express $\sigma(q^{k})$ in terms of $\sigma(q^{k-1})$ as

$$\sigma(q^{k}) = q\sigma(q^{k-1}) + 1 \tag{5}\label{eq5}$$

As you stated, suppose

$$\sigma(q^{k-1}) = a^2 \tag{6}\label{eq6}$$

$$\frac{\sigma(q^k)}{2} = b^2 \iff \sigma(q^k) = 2b^2 \tag{7}\label{eq7}$$

Substituting \eqref{eq6} and \eqref{eq7} into \eqref{eq5} gives

$$2b^2 = qa^2 + 1 \iff 2b^2 - qa^2 = 1 \iff (2b)^2 - (2q)a^2 = 2 \tag{8}\label{eq8}$$

Wikipedia's Pell's equation page's Transformations section gives a related equation of

$$u^{2}-dv^{2}=\pm 2 \tag{9}\label{eq9}$$

and how it can be transformed into the Pell equation form of

$$(u^{2}\mp 1)^{2}-d(uv)^{2}=1 \tag{10}\label{eq10}$$

Here, $u = 2b$, $v = a$, $d = 2q$ and the right side of \eqref{eq8} is $2$, so \eqref{eq10} becomes

$$((2b)^2 - 1)^2 - (2q)(2ba)^2 = 1 \tag{11}\label{eq11}$$

This is in Pell's equation form of $x^2 - ny^2 = 1$. Since $n = 2q$ is not a perfect square, there are infinitely many integer solutions. However, among these solutions, you first need to check that $x$ is in the form $4b^2 - 1$, the determined $b$ divides $y = 2ba$ and then that $a$ and $b$ satisfy \eqref{eq6} and \eqref{eq7} for some $k \equiv 1 \pmod 4$.

As for your open conjecture, if $k = 1$, then isn't $\sigma(q^{k-1}) = \sigma(q^{0}) = 1$ and $\frac{\sigma(q^{k})}{2} = \frac{\sigma(q)}{2} = \frac{1 + q}{2}$, so having both of them be squares requires $q = 2b^2 - 1$ for some $b$ and, thus, is not always true for all primes $q \equiv 1 \mod 4$, e.g., for $q = 5$, you get $5 = 2b^2 - 1 \implies 6 = 2b^2 \implies b = \sqrt{3}$?

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  • $\begingroup$ Thank you for your answer. Regarding your assertion in the last paragraph of your answer: Does the restriction $q = 2b^2 - 1$ for some $b$ lead to a contradiction? I doubt that it does, @JohnOmielan. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 6 at 13:38
  • $\begingroup$ @JoseArnaldoBebita-Dris I've given a very simple example where it doesn't work for all primes $q \equiv 1 \pmod 4$, i.e., $q = 5$. Am I misunderstanding something? $\endgroup$ – John Omielan Jul 6 at 16:21
  • $\begingroup$ Okay, I get your point @JohnOmielan. See zongxiangyi's proof for the implication $$\sigma(q^k)/2 \text{ is a square} \implies k = 1$$ and my own proof for the equivalence $$\sigma(q^{k-1}) \text{ is a square} \iff k = 1.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 6 at 16:24
  • $\begingroup$ I hereby retract my claim in the previous comment. Work is currently underway to prove the implication $$\sigma(q^{k-1}) \text{ is a square} \implies k = 1.$$ (The proof of the converse is trivial.) $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 6 at 21:07
  • $\begingroup$ Gladly accepting your answer now, @JohnOmielan. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 30 at 19:28
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(This proof is currently under reconstruction.)

Let $q$ be a prime satisfying $q \equiv k \equiv 1 \pmod 4$.

I (attempt to) prove here that

$$\sigma(q^{k-1}) = s(q^k) \text{ is a square } \implies k = 1.$$

Proof

Assume to the contrary that $k > 1$. This implies that $k \geq 5$ (since $k \equiv 1 \pmod 4$).

Suppose that $$s(q^k) = s^2 = \sigma(q^k) - q^k = \sigma(q^{k-1}) = \frac{q^k - 1}{q - 1}.\tag{$*$}$$

$(*)$ implies that $(q-1)s^2 = q^k - 1$, which is equivalent to $$q(q^{k-1} - s^2) = q^k - qs^2 = 1 - s^2 = (1 + s)(1 - s) = -(s+1)(s-1).$$

Since $q$ is prime, we consider three two cases:

Case 1: $q \mid s + 1$

SubCase 1.1: $q = s + 1$ $$\implies q - 1 = s \implies q^3 - 3q^2 + 3q - 1 = (q - 1)^3 = s^3 = (q - 1)s^2 = q^k - 1$$ $$\implies q^2 - 3q + 3 = q^{k-1} \geq q^4$$ This last inequality is a contradiction.

SubCase 1.2: $q < s + 1$

Take $1 < r = (s+1)/q$. Then from the equation $$q(s^2 - q^{k-1}) = (s+1)(s-1)$$ one gets $$s^2 - q^{k-1} = r(s - 1)$$ so that $$(s - 1) \mid (s^2 - q^{k-1}) = \sigma(q^{k-2})$$ where $s - 1 = \sqrt{\sigma(q^{k-1})} - 1$.
This implies that $(s - 1) \nmid q^{k-1}$ since $(s - 1) \mid (s^2 - q^{k-1})$ and $\gcd(s-1,s)=1$. In particular, $(s - 1) \nmid q^{k-1}$ implies that $$s \notin \left\{2, q+1, \ldots, q^{k-1} + 1\right\},$$ since the only possible divisors of $q^{k-1}$ are $1, q, \ldots, q^{k-1}$. But $q \mid (s+1)$. (No contradictions thus far.)

Note that $$\sigma(q^{k-2}) \equiv 1 + (k - 2) \equiv k - 1 \equiv 0 \pmod 4.$$ Also, we have the inequality $$\sqrt{\sigma(q^{k-1})} - 1 = s - 1 < s^2 - q^{k-1} = \sigma(q^{k-2}).$$ This last inequality implies that $$\sqrt{\frac{q^k - 1}{q - 1}} < \frac{q^{k-1} - 1}{q - 1} + 1 = \frac{q^{k-1} + q - 2}{q - 1}$$ from which we get $$\frac{q^k - 1}{q - 1} < \bigg(\frac{q^{k-1} + q - 2}{q - 1}\bigg)^2$$ which means that $$(q^k - 1)(q - 1) < (q^{k-1} + q - 2)^2.$$ (I am currently unable to get a contradiction under this SubCase 1.2.)

Case 2: $q \mid s - 1$

SubCase 2.1: $q = s - 1$ $$\implies q + 1 = s \implies q^{k-1} = s^2 - s - 1 = (q+1)^2 - (q+1) - 1$$ $$= q^2 + 2q + 1 - q - 1 - 1 = q^2 + q - 1$$ $$\implies q^2 + q - 1 = q^{k-1} \geq q^4$$ Again, this last inequality is a contradiction.

SubCase 2.2: $q < s - 1$

Take $1 < t = (s-1)/q$. Then from the equation $$q(s^2 - q^{k-1}) = (s+1)(s-1)$$ one gets $$s^2 - q^{k-1} = t(s + 1)$$ so that $$(s + 1) \mid (s^2 - q^{k-1}) = \sigma(q^{k-2})$$ where $s + 1 = \sqrt{\sigma(q^{k-1})} + 1$.

This implies that $(s + 1) \nmid q^{k-1}$ since $(s + 1) \mid (s^2 - q^{k-1})$ and $\gcd(s,s+1)=1$. In particular, $(s + 1) \nmid q^{k-1}$ implies that $$s \notin \left\{q-1, \ldots, q^{k-1} - 1\right\},$$ since the only possible divisors of $q^{k-1}$ are $1, q, \ldots, q^{k-1}$. But $q \mid (s-1)$. (No contradictions thus far.)

Note that $$\sigma(q^{k-2}) \equiv 1 + (k - 2) \equiv k - 1 \equiv 0 \pmod 4.$$ Also, we have the inequality $$\sqrt{\sigma(q^{k-1})} + 1 = s + 1 < s^2 - q^{k-1} = \sigma(q^{k-2}).$$ This last inequality implies that $$\sqrt{\frac{q^k - 1}{q - 1}} < \frac{q^{k-1} - 1}{q - 1} - 1 = \frac{q^{k-1} - q }{q - 1}$$ from which we get $$\frac{q^k - 1}{q - 1} < \bigg(\frac{q^{k-1} - q}{q - 1}\bigg)^2$$ which means that $$(q^k - 1)(q - 1) < (q^{k-1} - q)^2.$$ (I am currently unable to get a contradiction under this SubCase 2.2.)

QED

In fact, more is true.

If $k=1$, then $s(q^k) \text{ is a square}$.

Therefore, we have the biconditional $$s(q^k) = \sigma(q^{k-1}) = \frac{q^k - 1}{q - 1}$$ is a square if and only if $k=1$.

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  • $\begingroup$ The proof can be simplified by observing that $s(q^k) = \sigma(q^{k-1}) = s^2 \equiv 1 + (k-1) \equiv k \equiv 1 \pmod 4$ implies that $s$ is odd. The three cases (i) $q = s + 1$, (ii) $q = s - 1$, and (iii) $q = (s+1)(s-1)$ then all contradict $q \equiv 1 \pmod 4$. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 6 at 15:46
  • $\begingroup$ Please see the updated answer, which is currently under (re)construction. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 6 at 22:39

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