View Single Post
2011-05-31, 11:24   #5
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by CyD I didn't try to prove that any divisor of $2^{2^{n}}+1$ is like $X.2^{n+2}+1$. I know it's known. I used it in order to demonstrate the following (with the same notation than my previous message) $2^{2^{n}-i.(m+2)} = - (-X_m)^i mod D_{m,1}$ and for example, if $2^{n} = 0 mod (m+2)$ and with $i_{max} = \frac{2^{n}}{m+2}$ then $(-X_m)^{i_{max}} = -1 mod D_{m,1}$ and if you have already prove it and if you know some internet site or book, I am interested by that. Cyril
I will quote Serge Lang.