Thread: Proth theorem extended View Single Post
2009-04-07, 13:25   #5
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

2×7×103 Posts

Quote:
 Originally Posted by retina So where is the error in the "proof"? I think it would be quite instructive to see where it goes wrong.
Yes, this is far from a standard proof in math. Here it is a big mistake:
"now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^(2^(n-2)) == +/-1(mod p) implies that p == +/-1 (mod 2^n)"
This is totally false.
And R=Q in the "proof", if you haven't observed it.

Last fiddled with by R. Gerbicz on 2009-04-07 at 13:29