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Old 2011-05-31, 11:24   #5
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by CyD View Post
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.

Your notation sucks.

I can't be bothered wading through it. If you clean it up and repost
your comments, I will take a look.

Note, however, that trivially m+2 itself is a power of 2.
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