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 2022-12-19, 23:36 #1 Andrew Usher   Dec 2022 26·7 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 ...
 2022-12-20, 12:17 #2 alpertron     Aug 2002 Buenos Aires, Argentina 27628 Posts The number 2350896688821832838803657 is a factor of 2191 - 1 and it is congruent to 1 mod 1912.
 2022-12-20, 14:08 #3 Andrew Usher   Dec 2022 26×7 Posts (see next) Last fiddled with by Andrew Usher on 2022-12-20 at 14:20
2022-12-20, 14:19   #4
Dobri

"ม้าไฟ"
May 2018

2·5·53 Posts

Quote:
 Originally Posted by alpertron The number 2350896688821832838803657 is a factor of 2191 - 1 and it is congruent to 1 mod 1912.
However, 2350896688821832838803657 = 7068569257 × 332584516519201 is a composite factor.
I assume that the OP meant prime factors 1 mod p2 even though it was not specifically stated in the title of this thread.

 2022-12-20, 14:21 #5 Andrew Usher   Dec 2022 26·7 Posts That's correct.
 2022-12-20, 14:22 #6 kruoli     "Oliver" Sep 2017 Porta Westfalica, DE 23·3·5·13 Posts Still, both prime factors are 1 mod 191. Stupid, of course. Last fiddled with by kruoli on 2022-12-20 at 14:33 Reason: Redacted the obvious.
2022-12-20, 14:22   #7
alpertron

Aug 2002
Buenos Aires, Argentina

2×761 Posts

Quote:
 Originally Posted by Dobri However, 2350896688821832838803657 = 7068569257 × 332584516519201 is a composite factor. I assume that the OP meant prime factors 1 mod p2 even though it was not specifically stated in the title of this thread.
Well, that was not stated. I answered what was asked in the first post.

2022-12-20, 14:42   #8
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47·79 Posts

Quote:
 Originally Posted by Andrew Usher 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 ...
M1093 == 1 mod 1093^2 (M1093 is a composite factor of M1093 itself), the same holds for M3511

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^p-1)/(b-1) for various bases b (some bases b such as 47 and 72, have no single known Wieferich prime)

Last fiddled with by sweety439 on 2022-12-20 at 14:46

 2022-12-20, 21:18 #9 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 22·5·7·17 Posts I don't believe that any Mersenne number 2^p-1 with prime exponent p can have a prime factor q such that: valuation(q-1,p)>1 More generally numbers a^(p^n)-a^(p^(n-1)) for prime p and integers a and n can only have prime factors q such that: valuation(q-1,p) <= n-1 Last fiddled with by a1call on 2022-12-20 at 21:33
2022-12-20, 21:43   #10
alpertron

Aug 2002
Buenos Aires, Argentina

2×761 Posts

Quote:
 Originally Posted by Dobri However, 2350896688821832838803657 = 7068569257 × 332584516519201 is a composite factor. I assume that the OP meant prime factors 1 mod p2 even though it was not specifically stated in the title of this thread.
Let's try again:

The prime number 11686604129694847 is a divisor of 293077 - 1 and it is congruent to 1 mod 930772

2022-12-20, 21:47   #11
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

1001010011002 Posts

Quote:
 Originally Posted by alpertron Let's try again: The prime number 11686604129694847 is a divisor of 293077 - 1 and it is congruent to 1 mod 930772
I stand corrected.

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