The factorization of 2^10+1 has 1 factor of the form 10x+1: 5 * 5 * 41
The factorization of 2^100+1 has several factors of the form 100x+1 (i.e., ending in "01"): 17 * 401 * 61681 * 340801 * 2787601 * 3173389601
The recent factorization of 2^1000+1 has several factors of the form 1000x+1 (i.e., ending in "001"): 257 * 1601 * 25601 * 76001 * 42144001 * 82471201 * 4278255361 * 293543676001 * 432363203127002885506543172618401 * 4728377214827208130517802311270133709042152454873053138954405100001 * 1499880353767194377609094494397700575495188415483528578229421025616319442334213511407849468420747202400448847136686072102119598082440903794953849388001
There are also other visible patterns (5, 17, 257 are 2^(2^n)+1) (401, 1601, 25601). What is going on? Probably a simple extension of why Fermat numbers have to be of a particular form. Possibly related to 3^(3^(10^n))
Cunningham project
1 comment :
I found the following execises to the VI chapter of
The Number theory basics by Vinogradov, Moscow, Nauka, 1972
(sorry, I have only russian edition):
1b. Let a be an integer >1. Prove an odd prime divisor of a^p+1 divides a+1 or
is of form 2*p*x+1;
1d. Let n be an integer >0. Prove a prime divisor of 2^(2^n)+1 is of
form 2^(n+1)*x+1.
p - any odd prime.
aleksisto
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