Last night, I wrote:


   Does anyone know if there's a prime, n, such that:

   2^(n - 1) - 1 = 0 (mod n^2)

   I have convinced myself there aren't any such, but I can't prove it
   (yet?).


Well, I already have three responses indicating that two such primes
are 1093 and 3511 but that there are no others less than either 4 or 6
billion (depending on which reply I look at).  They are named
'Wieferich primes' after someone who asked the same question in 1909,
it seems.

On the other hand, neither of these primes divides a Mersenne of prime
exponent, so the square free conjecture is still undecided.
If s is the smallest natural number such that M(s) = 0 (mod n) and
n = k*s + 1 (note the lack of a '2*'), then:


n       s       M(s) % (n^2)    (n - 1)         s               k

1093    364     0               2^2*3*7*13      2^2*7*13        3
3511    1755    0               2*3^3*5*13      3^3*5*13        2

('%' is the C programming language's integer modulo operator in case
someone out there doesn't know).

In both cases, M(s) % (n^3) is not zero.

Also, note that the table contains a trend that would indicate that
there are no other Wieferich primes: (n - 1) cannot lose another
factor of 2 if n is prime.  Further, if k goes to 1, then (n - 1) = s,
making the exponent composite again.  But, of course, generalizing
from two cases is a bad habit I should try to break.:)

Lastly, if (n - 1) can be shown to always have only small factors if n
is a Wieferich prime, then the square free conjecture can be proven
that way.

Many thanks to those that replied (or were about to :),


                                                Will