20221219, 23:36  #1 
Dec 2022
313 Posts 
Are any Mersenne factors 1 mod p^2?
By heuristics there should be infinitely many, but I wonder whether there's not a prohibition similar to the Wieferich condition at work here. As with that kp^2+1 divides phi(p) at infinitely many values provably, but 2 doesn't seem to be one of them ...

20221220, 12:17  #2 
Aug 2002
Buenos Aires, Argentina
1500_{10} Posts 
The number 2350896688821832838803657 is a factor of 2^{191}  1 and it is congruent to 1 mod 191^{2}.

20221220, 14:08  #3 
Dec 2022
313 Posts 
(see next)
Last fiddled with by Andrew Usher on 20221220 at 14:20 
20221220, 14:19  #4  
"ม้าไฟ"
May 2018
2^{4}·29 Posts 
Quote:
I assume that the OP meant prime factors 1 mod p^{2} even though it was not specifically stated in the title of this thread. 

20221220, 14:21  #5 
Dec 2022
100111001_{2} Posts 
That's correct.

20221220, 14:22  #6 
"Oliver"
Sep 2017
Porta Westfalica, DE
1,447 Posts 
Stupid, of course. Last fiddled with by kruoli on 20221220 at 14:33 Reason: Redacted the obvious. 
20221220, 14:22  #7 
Aug 2002
Buenos Aires, Argentina
2^{2}·3·5^{3} Posts 
Well, that was not stated. I answered what was asked in the first post.

20221220, 14:42  #8  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×7×263 Posts 
Quote:
For the case of decimal repunit, there is a small prime solution: 37 divides R3 (= 111) and 37 == 1 mod 3^2 You can also consider the case of Wagstaff numbers (2^p+1)/3 and the generalized repunits (b^p1)/(b1) for various bases b (some bases b such as 47 and 72, have no single known Wieferich prime) Last fiddled with by sweety439 on 20221220 at 14:46 

20221220, 21:18  #9 
"Rashid Naimi"
Oct 2015
Remote to Here/There
100100111111_{2} Posts 
I don't believe that any Mersenne number 2^p1 with prime exponent p can have a prime factor q such that:
valuation(q1,p)>1 More generally numbers a^(p^n)a^(p^(n1)) for prime p and integers a and n can only have prime factors q such that: valuation(q1,p) <= n1 Last fiddled with by a1call on 20221220 at 21:33 
20221220, 21:43  #10  
Aug 2002
Buenos Aires, Argentina
2^{2}×3×5^{3} Posts 
Quote:
The prime number 11686604129694847 is a divisor of 2^{93077}  1 and it is congruent to 1 mod 93077^{2} 

20221220, 21:47  #11 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2367_{10} Posts 

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