Hello!
I was fiddling around with Mersenne factors the other day and came across
an interesting result. I'm not sure if it is of any use, but I think if
anyone can extend it just a little further, then it would cut down the
numbers we would have to try to factor by a HUGE amount...
Anyway, any mersenne's factor can be written as 2kp+1
So any persenne number 2^p - 1 (here on written as P) can be written as
(2ap+1)(2bp+1) = P i.f.f. it is not a prime
Now define x = p(a+b)+1, y=p(a-b)
Then x+y=2ap+1, x-y=2bp+1
so P=(x-y)(x+y)
x^2 - y^2 = P
Now write n=a+b, m=a-b so x=pn+1 , y=pm
Then (pn)^2 + 2pn + 1 + (pm)^2 = P
taking this mod p^2 and rearranging a little
2pn = P-1 mod p^2
this means 2pn = (P-1) + pZ for some integer Z
so (P-1) must have a factor of p, which we can cancel (we also know this
directly). Call (P-1)/p = Q
Then 2n = Q mod p
n = Q/2 mod p which is well defined
Therefore we can find the sun of the two factors mod p.
I have tried (and failed so far) to get a similar relationship for y (or a
or b)
If we could get such a relationship for y, and we assumed we were looking
for the smallest factor, then we could work out something about that
factor (hopefully it's value mod p) which would cut down on factoring
requirements.
Any ideas?
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Chris Jefferson, Girton College, Cambridge, [EMAIL PROTECTED]
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Someone may have beaten me to Fermat's Last Theorem, but I've done
Riemann's general equation. However it won't fit in my signature file...
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