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Old 2015-09-08, 19:38   #5
Drdmitry's Avatar
Nov 2011

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Originally Posted by R.D. Silverman View Post
You claimed to KNOW how to do it for quadratics. Do you withdraw this claim?

I certainly do not know how to do it.
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:
P(1240169989195728649392349829489369369557206090108450826526311603480495971405669477171430401412715470) =
844293027521791885331059004528757978188025863965373325668959095809803279961413930426485139159366291 *
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.

Last fiddled with by Drdmitry on 2015-09-08 at 19:40
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