Theorem with Exceptions

[Tomasz Ordowski]

Hello Math Fans! 

                                

If a prime p > 5 and 2^(p-1) =/= 1 (mod p^2),  

then lambda((2^(p-1)-1)/p) = lambda(2^(p-1)-1),

where lambda is the Carmichael function. 

Do the known Wieferich primes (1093 and 3511)
actually do not satisfy the equality in this theorem? 

Or maybe there are other exceptions?

I do not think so, see the proof in PS. 

Best, 

Tom Ordo 

______________

PS. Proof sketch: 

Put M = 2^(p-1) - 1. Since p is not Wieferich, p divides M exactly once. 

It therefore suffices to show that p - 1 still occurs in the Carmichael lcm after p is removed.

For p >= 17, consider the cyclotomic factor Phi_(p-1)(2). It is greater than p. Every prime q dividing this factor, except possibly q = p, satisfies ord_q(2) = p - 1, hence q = 1 (mod p - 1). Since p occurs only once in M, Phi_(p-1)(2) cannot consist solely of p. Thus some q <> p remains after division by p, and q - 1 is divisible by p - 1. Hence lambda(M/p) = lambda(M).

The cases p = 7,11,13 are checked directly. Thus the equality holds for every non-Wieferich prime p > 5.

This also explains why the known Wieferich primes p = 1093 and 3511 are the natural cases to test separately. 

_______Similar question at the end_____________

Are p = 3, 5, 7, 11, 13, 17, 41, and 73 the only exceptions to the identity

lambda((2^((p-1)/2) − (2/p))/p) = lambda(2^((p-1)/2) − (2/p)),

for odd primes p, where (2/p) is the Legendre symbol?