View Single Post
Old 2015-09-08, 21:06   #6
xilman
Bamboozled!
 
xilman's Avatar
 
"π’‰Ίπ’ŒŒπ’‡·π’†·π’€­"
May 2003
Down not across

277B16 Posts
Default

Quote:
Originally Posted by Drdmitry View Post
Yes, I do know how to do it.
Initially I programmed a generator for the polynomial P(x) = x^2 + 1 since it is the easiest case. However it is not too difficult to adapt it for P(x) = x^2 - x + 1. It generated the following factorisation:
Code:
P(1240169989195728649392349829489369369557206090108450826526311603480495971405669477171430401412715470) =
844293027521791885331059004528757978188025863965373325668959095809803279961413930426485139159366291 *
1821668013315479067495522348143574102954344772015219924580113016806982758457314260330321273331462541
One prime divisor has 99 digits and another one is 100 digits long.

P.S. Just realised that there should be one more condition on P(x): for each prime q there exists an integer a such that P(a) is not zero modulo q.
Interesting. Please describe your algorithm in greater detail.
xilman is offline   Reply With Quote