# [seqfan] An unresolved problem

Tomasz Ordowski tomaszordowski at gmail.com
Sun Oct 21 07:15:19 CEST 2018

```Dear SeqFans!

As is well known:

If p is a prime that does not divide b-1, then (b^p-1)/(b-1) has all
divisors d == 1 (mod p).

Let a(n) be the smallest composite k such that (n^k-1)/(n-1) has all
divisors d == 1 (mod k).

Note that (n^k-1)/(n-1) == 1 (mod k), so n^k == n (mod (n-1)k). Thus a(n)
>= A298908(n).

Cf. https://oeis.org/history/view?seq=A298908&v=25

a(n) is the least composite k such that (n^k-1)/(n-1) has all prime
divisors p == 1 (mod k).

For n > 2, a(n) = 91, 4, 5731, 6, 25, 9, 91, 10, ... Please more.

The basic problem: a(2) = ?

The question: Are there pseudoprimes k = pq such that both 2^p-1 and 2^q-1
are prime?

Theorem: If k is such a pseudoprime, then 2^k-1 has all divisors d == 1
(mod k).

Cf. https://oeis.org/A298076

Can also consider these numbers to negative bases by defining:

Least odd composite k such that (n^k+1)/(n+1) has all divisors d == 1 (mod
k).

Data: 9, 341, 91, ... for n = 1, 2, 3, ... Cf. https://oeis.org/A298310

The most interesting is the base -2, so let's define:

Odd composite numbers k such that (2^k+1)/3 has all divisors d == 1 (mod
k).

I found only two such numbers k = 341 = 11*31 and k = 5461 = 43*127. Please
more.

If both (2^p+1)/3 and (2^q+1)/3 are prime and n = pq is pseudoprime, then n
is such a number k.

Do known the Wagstaf exponents https://oeis.org/A000978 give more such
pseudoprimes?

Does it exhaust all other cases?

Best regards,

Thomas
____________________
The numeric coincidence:
These two numbers 341 and 5461 are also Cipolla pseudoprimes:
https://oeis.org/A210454

```