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.


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 $q<n$ 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 $n<q^k$ contradicts the Dris Conjecture.


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

  • $\begingroup$ What does special prime means? $\endgroup$ – Aqua Feb 21 at 16:13
  • $\begingroup$ @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 $\gcd(q,n)=1$.) $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 21 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.

  • $\begingroup$ 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 $$ \operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\operatorname{gcd}(q^{k+1}-1,(q^k+1)^2)? $$ $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 22 at 6:25
  • $\begingroup$ 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 $$\operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) = q^{k+1} - 1.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 22 at 6:33
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    $\begingroup$ Not a typo, I'm just using $\operatorname{gcd}(a,bc) \le \operatorname{gcd}(a,b)\operatorname{gcd}(a,c)$. $\endgroup$ – FredH Feb 22 at 7:28
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    $\begingroup$ If $(*)$ holds, $q^{k+1}-1$ is a divisor of the RHS, so their GCD is $q^{k+1}-1$. $\endgroup$ – FredH Feb 22 at 7:29
  • $\begingroup$ Thank you very much for your follow-up comments, @FredH! I think I get it now. =) $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 22 at 7:39

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