[seqfan] Converse

[Tomasz Ordowski]

Hello!

Let us define:

Numbers n such that 2^n == 2 (mod n).

Theorem: if n is such a number,
    then 2^n-1 is such a number.

Problem: is the reverse theorem true?
 The converse holds up to n = 10000.

Generally:

Let B be the set of natural numbers n such that

         (b^n-1)/(b-1) == 1 (mod n)

with a fixed integer b > 1.

Conjecture:

The number (b^k-1)/(b-1) belongs to B if and only if k belongs to B.

Note: This can be extended to bases b < -1.

Do you have an idea how to prove my conjecture?
Professor Pomerance does not know a proof.
Please reply, it's important to me!

Sincerely,

Thomas Ordowski
________________________
Steuerwald's theorem (1948):
If k is a psp(b) and gcd(k,b-1)=1, then (b^k-1)/(b-1) is a psp(b).
Here a composite k is a psp(b) if and only if b^k == b (mod k).