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2015-09-08, 21:06   #6
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

10,243 Posts

Quote:
 Originally Posted by Drdmitry 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.