0

The topic of odd perfect numbers likely needs no introduction.

Let $N$ be an odd perfect number given in the so-called Eulerian form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

It is known that $n^2 - q^k$ is not a square. (Note that $n^2 - q^k \equiv 0 \pmod 4$ is true since $n$ odd implies that $n^2 \equiv 1 \pmod 4$ holds. In particular, we may write $n^2 - q^k = 2^r t$, where $\gcd(2,t)=1$ since $n^2 - q^k$ is not squarefree.)

Since $n^2 - q^k = 2^r t$ is not a square and $\gcd(2,t)=1$, then $3 \leq r$ is odd. Therefore, since $n$ is odd implies that $n^2 \equiv 1 \pmod 8$ holds, and since $2^r t \equiv 0 \pmod 8$, it follows that $q^k \equiv 1 \pmod 8$. Hence, we infer that $q \equiv 1 \pmod 8$ holds, by considering the various possibilities for $q \pmod 8$ and $k \pmod 8$, where we note that $q$ and $k$ are both odd. (Furthermore, they are both congruent to $1 \pmod 4$, but that will not matter in the argument.) $\tag{1}$

Consequently, $q \equiv 1 \pmod 8$ rules out $q = 5$ and $q = 13$. We therefore conclude that $q \geq 17$.

Here is my:

QUESTION: I have doubts about the logical validity of the argument in the section marked with a $(1)$. Is the proof logically sound? If it is not, how can it be mended so as to produce a correct proof?

I would have included a table showing all of the possibilities for $q \pmod 8$ and $k \pmod 8$, and computing $q^k$ for each such possibility, but I have forgotten how to typeset tables using MathJAX. I hope this is OK.

Updated - January 22, 2023 - 18:04 PM (Manila time)

While the argument in the original post is flawed, we do get $3 \leq r$ under the additional hypothesis that $\sigma(n^2)/q^k$ is a square. (Please see [Dris and San Diego (2020), pages 31-32].)

Bill Dubuque
  • 282,220
Jose Arnaldo Bebita Dris
  • 8,471

2 Answers2

1

Sounds sound enough, but the fact $n^2 \equiv 1 \pmod8$ sounds a little confusing in my opinion, so putting the fact out that $q^k\equiv 1\pmod8$ should come first, and then $2^rt \equiv 0\pmod 8$, and then $n^2 \equiv 1 \pmod8$.

However, take all of this advice with a grain of salt as it is with personal opinion.

A Neutrino Boy
  • 73
  • 7
  • Thank you for your time and attention, @ANeutrinoBoy! The fact that $q^k \equiv 1 \pmod 8$ is derived from $2^r t \equiv 0 \pmod 8$ and $n^2 \equiv 1 \pmod 8$. The fact that $n^2 \equiv 1 \pmod 8$ follows from the fact that $n$ is odd - indeed, it suffices to consider the residue classes $n \equiv 1 \pmod 8$, $n \equiv 3 \pmod 8$, $n \equiv 5 \pmod 8$, and $n \equiv 7 \pmod 8$. – Jose Arnaldo Bebita Dris Dec 28 '22 at 10:41
1

I don't think that you can conclude that "$3\leq r$ is odd".

For $r=2$ and any non-square odd integer $t$, $n^2-q^k=2^{r}t$ is not a square.

By the way, one can prove that $q^k\equiv 1\pmod 8$ implies $q\equiv 1\pmod 8$ as follows : $$1\equiv q^k\equiv q\cdot (q^2)^{\frac{k-1}{2}}\equiv q\cdot 1^{\frac{k-1}{2}}\equiv q\pmod 8$$

mathlove
  • 151,597