# If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.

Problem Statement

Prove the following proposition.

If $$q^k n^2$$ is an odd perfect number with special prime $$q$$, then $$n^2 - q^k$$ is not a square.

Motivation

Let $$q^k n^2$$ be an odd perfect number with special prime $$q$$. Then $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$.

By Pomerance, et al., we know that $$q^k < n^2$$, so that $$n^2 - q^k$$ is a positive integer. Also, since $$n^2$$ is a square and $$q \equiv 1 \pmod 4$$, then $$n^2 - q^k \equiv 1 - 1 \equiv 0 \pmod 4.$$

My Attempt

Suppose that $$q^k n^2$$ is an odd perfect number with special prime $$q$$, and that $$n^2 - q^k = s^2$$, for some $$s \geq 2$$.

Then $$n^2 - s^2 = q^k = (n + s)(n - s)$$ so that we obtain $$\begin{cases} {q^{k-v} = n + s \\ q^v = n - s} \end{cases}$$ where $$v$$ is a positive integer satisfying $$0 \leq v \leq (k-1)/2$$. It follows that we have the system $$\begin{cases} {q^{k-v} + q^v = q^v (q^{k-2v} + 1) = 2n \\ q^{k-v} - q^v = q^v (q^{k-2v} - 1) = 2s} \end{cases}$$

Since $$q$$ is a prime satisfying $$q \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$, from the first equation it follows that $$v=0$$, so that we obtain $$\begin{cases} {q^k + 1 = 2n \\ q^k - 1 = 2s} \end{cases}$$ which yields $$n = \frac{q^k + 1}{2} < q^k.$$ Lastly, note that the inequality $$q has been proved by Brown (2016), Dris (2017), and Starni (2018), so that we are faced with the inequality $$q < n < q^k.$$ This implies that $$k>1$$.

Finally, notice that $$k>1$$ contradicts the Descartes-Frenicle-Sorli Conjecture, while $$n contradicts the Dris Conjecture.

Question

Is it possible to remove the reliance of this proof on the truth of either the Descartes-Frenicle-Sorli Conjecture or the Dris Conjecture?

• What does special prime means? Feb 21, 2019 at 16:13
• @greedoid, the special prime $q$ is also called the Euler prime. It is the prime divisor which occurs to an odd exponent in the factorization of the odd perfect number $q^k n^2$. (That is, we have $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.) Feb 21, 2019 at 16:16

Here's a way to finish the proof without appealing to any conjecture.

If $$q^k n^2$$ is a perfect number with $$\operatorname{gcd}(q,n)=1$$, we have $$\sigma(q^k) \sigma(n^2) = 2 q^k n^2.$$ We know that $$\sigma(q^k) = (q^{k+1}-1)/(q-1)$$ and you've shown that $$n = (q^k + 1)/2$$, so we can conclude that $$2(q^{k+1}-1) \sigma(n^2) = (q-1) q^k (q^k + 1)^2.\tag{*}$$ Consider the GCD of $$q^{k+1}-1$$ with the right-hand side: $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\operatorname{gcd}(q^{k+1}-1,q^k+1)^2,$$ since $$q^k$$ is coprime to $$q^{k+1} - 1$$.

Noticing that $$q^{k+1} - 1$$ = $$q(q^k + 1) - (q + 1)$$, we find $$\operatorname{gcd}(q^{k+1}-1,q^k+1) = \operatorname{gcd}(q+1,q^k+1)$$, which is $$q+1$$ because $$k$$ is odd.

Thus $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)(q+1)^2.$$ Since $$k\equiv 1 \pmod 4$$ and you have shown $$k \gt 1$$, we have $$k \ge 5$$. If $$(*)$$ holds, the left-hand side of the inequality must be $$q^{k+1}-1$$, which is then greater than $$q^5$$. But the right-hand side is less than $$q^4$$, so this is impossible.

• Very nice, @FredH! I think there is a typo in $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\operatorname{gcd}(q^{k+1}-1,q^k+1)^2,$$ shouldn't that be $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\operatorname{gcd}(q^{k+1}-1,(q^k+1)^2)?$$ Feb 22, 2019 at 6:25
• Also, can you comment more on why the left-hand side of the inequality $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)(q+1)^2$$ must be $q^{k+1} - 1$? That is, it appears that you are claiming that $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) = q^{k+1} - 1.$$ Feb 22, 2019 at 6:33
• Not a typo, I'm just using $\operatorname{gcd}(a,bc) \le \operatorname{gcd}(a,b)\operatorname{gcd}(a,c)$. Feb 22, 2019 at 7:28
• If $(*)$ holds, $q^{k+1}-1$ is a divisor of the RHS, so their GCD is $q^{k+1}-1$. Feb 22, 2019 at 7:29
• Thank you very much for your follow-up comments, @FredH! I think I get it now. =) Feb 22, 2019 at 7:39