In the early days of Covid-19 I began a renewed study of Mersenne numbers (prime and composite) and a surprise was the critical role played by Wieferich primes (strictly base-2 Wieferich primes). Later my work branched out into generalised Mersenne numbers inspired by a 1967 paper of Le Roy J. Warren and Henry G. Bray concerning square-free Mersenne numbers (one in which base-2 Wieferich primes yet again played a critical role). Later again Sophie Germain primes entered the scene...

A longstanding conjecture concerning Mersenne numbers. Let p be prime and (Mersenne) M_p = 2^p - 1; then M_p is squarefree.

A particular case of Warren-Bray. Let p be prime and (Mersenne) M_p = 2^p - 1. Let q be a prime such that q²|M_p, then q is a (base-2) Wieferich prime [!! amazing!!]

Comment. Who would ever have guessed that?! When I first came upon that result I was filled with such disbelief, and my first impulse was to identify a flaw in the proof (I first came upon it here)... The proof is so short, and one cannot find an error, for it is as correct as anything could be. (Oh how I wish I hadn't seen the result and proof all together... Dear reader, here an exercise for you. Without looking up the proof, find a proof for yourself. It's the sort of thing that if you're told the result is (genuinely) true, then you will have no difficulty making a proof of your own.) Incidentally, I easily extended the proof to the numbers I deemed to be generalised Fermat-Mersenne numbers in the following unpublished paper of mine Fermat_6.

Aside. Someone who read that unpublished paper of mine attributed those generalised numbers to me at the Wikipedia page on Mersenne primes, see here. Scroll down to the section with title Mersenne-Fermat primes.

They gave a link to a now vanished early website of mine, and I don't know how to edit a Wikipedia page to update the link. For the moment the referenced page is available at the WaybackMachine, last archived in 2016 here. In the unlikely event that someone reads this - who also knows how to edit a Wikipedia page - would they please write to inform me what to do. (Oh, I might ask AI Claude...)

I am not interested in attempting to publish any of this (too extensive) work - work whose complete details no one would ever have enough time to read and verify all the details - but one day I will give a brief summary of it here. I believe that anyone who will read it with an open mind will see there is something new and non-trivial in it.

In the meantime I'll just give a brief glance at what I intend writing about:

I begin with a well-known general observation: let G is any group (not necessarily abelian) and p be a prime divisor of |G| (the order of G), then G contains an element of order p. However, if p² divides |G|, it isn't necessarily true that G contains an element of order p² (even in the abelian case).

The work I present here didn't start with that observation, but I have chosen to begin with it, and pose the following simple question of a very specific group G, namely the multiplicative group of residues modulo M_p, where M_p is the Mersenne number M_p = 2^p - 1, p prime.

Watch this space!

Working note for myself (from my Fermat6 page). html code for Mersenne and Fermat numbers, and more:
The Mersenne numbers are the M_p = 2^p - 1, p prime; the Fermat numbers are the F_n = 2²ⁿ + 1, n = 0, 1, 2, 3, 4, 5, ... . What does this produce?: M_p(b) = (b^p - 1) / (b - 1).
This is the code for A/B: $\frac{A}{B}$
Try this out: $\frac{\mathrm{(<i>b</i>^<i>p</i>\; -\; 1)}}{\mathrm{(<i>b</i>\; -\; 1)}}$
Remove the (brackets): $\frac{\mathrm{<i>b</i>^<i>p</i>\; -\; 1}}{\mathrm{<i>b</i>\; -\; 1}}$
And now this: M_p(b) = $\frac{\mathrm{<i>b</i>^<i>p</i>\; -\; 1}}{\mathrm{<i>b</i>\; -\; 1}}$
And this: M_p(b) = $\frac{\mathrm{<i>b</i>^<i>p</i>\; -\; 1}}{\mathrm{<i>b</i>\; -\; 1}}$
https://www.toptal.com/designers/htmlarrows/math/

This is the code for 'not equal to': ≠
Code for 'is congruent to': ≡
For not congruent: ≢
This: p² [the '2' on the big side]
This: p² [the '2' reduced in size]
This is the code for p-1/3: $\frac{\mathrm{p-1}}{3}$
Adapt to: This is the code for p-1/3: $\frac{3}{}$
Then to: This is the code for p-1/3: $\frac{}{}$3
No, try this: $\frac{}{}$p-1/3
Try this: ($\frac{\mathrm{<i>p</i>-1}}{3}$)!
Now this: ($\frac{\mathrm{<i>p</i>-1}}{3}$)! ≡ 1 (mod p)
What about this?: ($\frac{\mathrm{<i>p</i>-1}}{3}$)_n! ≡ 1 (mod p)
Gauss factorial of N w.r.t. n: N_n!

Wilson's theorem: (p - 1)! ≡ -1 (mod p)

Notation. For integers N and n we define (N - 1)_n! to be the product of all the integers in the intervan [1, N] that are relatively prime to n.

Gauss-Wilson theorem: (n - 1)_n! ≡ -1 (mod n) if n = p^a or n = 2.p^a, p an odd prime, integral a ...