I was curious about what is known about the density of the primes such that 2 has odd order modulo p. Is that density well-known? Is there more general answer to this question when we replace 2 with some other number? Thanks.
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See this paper:https://projecteuclid.org/download/pdf_1/euclid.pjm/1102706452 and my answer https://math.stackexchange.com/questions/3018887/about-the-parity-of-ord-p7/3020906#3020906 The density is well-known, it is $7/24$. – Sungjin Kim Dec 08 '18 at 18:41
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From quadratic reciprocity when $p\equiv 7\pmod 8$, $2$ would be a square mod $p$. In this case $(p-1)/2$ would be an odd number. Writing $2=a^2$ we see that by Fermat's Little theorem $(a^2)^{(p-1)/2}\equiv 1$. This means $2^{(p-1)/2}$ is $1$ mod $p$. So $2$ has odd order.
Dirichlets' theorem on arithmetic progressions gives the density of primes that are $7$ modulo 8. So the density you want is at least that much.
P Vanchinathan
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Thanks for the answer. What you're saying is definitely right, but I suspect that density of such primes might be strictly bigger than 1/4. When I thought about this, I tried couple of other primes that are not 7(mod 8), and there are really some that works. (e.g. 73. Order of 2 (mod 73) is 9.) I'm interested in the exact density, if it's somehow known in the literature at all. – vgmath Oct 31 '16 at 00:50
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Playing with a few examples I see that it might be true that
Conjecture: When $2^n-1$ is prime then $2^n-1$ is the only prime for which $2^n\equiv 1\pmod{2^n-1}$. If $2^n-1$ is composite, then for each prime $q\mid n$ we have $2^n\equiv 1\pmod{q}$.
If this is true, then these numbers have a density bounded below by the density of the Mersenne numbers.
I don't know if this is better or worse that the previous suggestion given by P Vanchinathan...or even if it is too trivial to be useful.
Laars Helenius
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