View Single Post
Old 2005-09-28, 17:53   #10
Yamato
 
Yamato's Avatar
 
Sep 2005
Berlin

2×3×11 Posts
Default

Let F(n) = 2^(2^n) + 1.

Then the "trivial" square root of 2 is

sqrt(2) = F(n-1) * inv(F(n-2) - 1) (mod F(n))
(inv is the inverse number mod F(n))

because of the real factorization

F(n) = F(n-1)^2 - x^2

where x = (F(n-2) - 1)*sqrt(2)
Yamato is offline   Reply With Quote