None of the Mersenne's numbers (2^n-1) where n is of the form 2^k has prime factors of the form 4k-1 (easy to demonstrate).

You can see below that all the others seem to have at least one (in red), but I can't prove it. There is a periodicity for some factors (like 7, 11, 17, 31, ...) but I don't see yet the general rule.

See the series of numbers in which some are verifying and some are NOT.

Ok but i need to find them all

Maybe some solutions aren't in this

It seems work if and only if n=2^k (with k integer), but I cannot prove it. I keep searching.

Hmmmm... maybe it is. i can't proove it :(

A possible approach :

- a²+9 is a sum of two squares

- a prime of the form 4k-1 divides the sum only if it divides both squares (Euler)

- the only prime of the form 4k-1 that divides 9 is 3

- so, the dividor 2^n-1 of your problem must have only prime factors of the form 4k+1 except 3.

All for now. I follow this track tonight.

None of the Mersenne's numbers (2^n-1) where n is of the form 2^k has prime factors of the form 4k-1 (easy to demonstrate).

You can see below that all the others seem to have at least one (in red), but I can't prove it. There is a periodicity for some factors (like 7, 11, 17, 31, ...) but I don't see yet the general rule.