Are any Mersenne factors 1 mod p^2?

[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 ...

[alpertron]

The number 2350896688821832838803657 is a factor of 2¹⁹¹ - 1 and it is congruent to 1 mod 191².

[Andrew Usher]

(see next)

[Dobri]

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.

[Andrew Usher]

That's correct.

[kruoli]

Still, both prime factors are 1 mod 191.
Stupid, of course.

[alpertron]

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.

[sweety439]

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)

[a1call]

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

[alpertron]

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

[a1call]

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.