Hi!
Just something I was pondering a couple of days ago...
Consider a general number (odd) number c which can be factored into ab=c
W.L.O.G. assume b is greater than a
then let x=(a+b)/2 , y=(b-a)/2
then (x+y)(x-y)=c
x^2 - y^2 = c
x^2 = c + y^2
So if we can find if this equation has any integer solutions, we've found
our factors...
Ways of doing this:
The difference of two squares is always an arthimetic progression of odd
numbers. Here is an example..
2^2 - 1^2 = 3
3^2 - 2^2 = 5
4^2 - 3^2 = 7
and so on... So look at general sum of an arthimetic series
S(n) = (n/2)(2a + (n-1)d) In this case d=2 and a is odd, so need to try to
solve c = na + n(n-1)/2 for integers n,a
Also, try to solve x^2 - y^2 = 0 mod c
As if this is solvable, then (x-y)(x+y)=nc, for integer n, so must be able
to cancel out all factors of n in either (x-y) or (x+y) to get back to a
solution of equation..
Alternativly, could try to find out by some kind of set notation what the
size of the group of solns. is... This is where I come unstuck.
I believe this is an example of an eliptic curve, and I want the c'th term
in it's L-series. Could we transform it into a modular form and then
quickly work out this term. I could well be in cloud-cockoo land now, as I
aren't even totally sure what a modular form is, but I know that the
L-series of modular forms, and some series related to modular forms are
the same, and this proof lead the the solution of Fermat's Last
Thereom....
Anyway, if anyone could just vaguely point me in the right direction, or
tell me if I am talking rubbish before I go and start reading up on all
this... Thanks!
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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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