Hmm... Thanks for all the advice on factoring!
Just one question / comment.
We want to find 2^p mod n where n is in form 2kp + 1
If we KNOW 2kp + 1 is prime, then the euler totient fn. of n is 2kp, call
this T
Also, if a is co-prime to n, a^T=1 mod n
2 is obviously co-prime to n, so 2^T=1 mod n
So another way of working out if a factor would be to work out
So if 2kp + 1 is prime, 2^(2kp)=1 mod (2kp + 1)
Herein lies my problem:
I know that if p is prime, 'mod p' is a group under muliplication.
I know that 2*(2^(p-2))=1 mod p
so 2^(p-2) is the inverse of 2 and only by multilplying by 2^(p-2) can I
get 1.
Now, this implies that for no 0<N<2kp, 2^N=1 mod (2kp + 1)
so 2^p cannot be 1 mod (2kp + 1)
?????
Sorry to bother you again
Chris Jefferson, Cambridge
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