# multiplicative order of 2 mod p

Can anyone tell me whether or not it is true that for all odd primes p the multiplicative order of 2 modulo p is strictly less than the multiplicative order of 2 modulo p^2 ? What are some good references regarding this problem ? Thank you

-
The first counterexample is $p=1093$. See en.wikipedia.org/wiki/Wieferich_prime – François Brunault Nov 10 '12 at 16:10
I am a bit surprised this gets ever more votes to close (four already). In my opinion this is an alright question. Could someone at least say why they want this to be closed. – user9072 Nov 10 '12 at 18:34
I think people voted against this question, because it is easily equivalent to a question that is discussed in introductory textbooks (e.g. Hardy-Wright, Ireland-Rosen, Nathanson). – GH from MO Nov 10 '12 at 19:33
Related to an old (rather feeble!) question of mine: mathoverflow.net/questions/27579/… – David Loeffler Nov 11 '12 at 9:10

It is well-known that there are primes $p$ such that $2^{p-1} \equiv 1$ (mod $p^{2}$), a question which arises in connection with Fermat's Last Theorem. For such a prime $p,$ let $e$ be the smallest positive integer such that $p$ divides $2^{e}-1,$ and write (as we may) $p-1 = ed$ with $d$ an integer. Then we see easily that $2^{p-1}-1 \equiv d(2^{e}-1)$ (mod $p^{2}$). Certainly $d$ is not divisible by $p,$ so we must already have $2^{e} \equiv 1$ (mod $p^{2}$). Hence for such a prime $p,$ the multiplicative order of $2$ (mod $p$) is the same as the multiplicative order of $2$ mod $p^{2}.$ I see in the meantime that Francois Brunault has made a comment to similar effect, and that $1093$ is the smallest such prime