you "just" proved that
3^(2^1048576) != 1 (mod 2^1048576 + 1)
then F20 is composite (Fermat theorem).
It doesn't help to find a factor.
I tried to find a factor of F22, using Pollard p-1 method and reached
B=2,000,000 without finding a divisor.
Does anyone know current limit of Pollard p-1 method for F20 ? Because
divisors of Fermat numbers minus one are divisible by 2^(n+2) a large
attempt with Pollard p-1 should be tried before using ECM !?
Yves
>hey all. i am very excited about finding a new math discovery.
>i wanted to find a factor of fermat's number 20. - F20
>so.. using Proth.. i set it up with
>Start: For n=1048576 to 1048576, For k=2 to 2 step 2, GFN.
>1048579 being 2^20.
>after 3 days of processing on my Intel 300 Mhz..
>it came out with this result in the log file..
>1*2^1048576 + 1 is composite. (a = 3)
>what is the relevance of a?
>and HOW can i use my computer to find a factor of F20?
>thank you.
>-paul
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