View Single Post
Old 2017-07-30, 09:21   #3
R. Gerbicz
 
R. Gerbicz's Avatar
 
"Robert Gerbicz"
Oct 2005
Hungary

26648 Posts
Default

Quote:
Originally Posted by carpetpool View Post
Now we see that the discriminant D of S(x, n) = 2^r where r = (n+1)*2^n-1
Prove it.
Quote:
Originally Posted by carpetpool View Post
Since the exponent r is odd, each prime factor q dividing S(x, n) has the form k*2^(n+1)+-1.
This is very false, it fails even for n=3: 2 divides S(3), but you can't write 2 in the k*2^4+-1 form.
R. Gerbicz is offline   Reply With Quote