mersenneforum.org How to find values of polynomials with nice factorization?
 Register FAQ Search Today's Posts Mark Forums Read

 2015-09-08, 15:11 #1 Drdmitry     Nov 2011 3528 Posts How to find values of polynomials with nice factorization? Given an irreducible polynomial P(x) with integer coefficients, is there any reasonable algorithm known, which constructs a random n such that P(n) is a product of two huge prime factors of relatively same size? I know how to do that for quadratic polynomials. But what about higher degrees?
2015-09-08, 15:47   #2
R.D. Silverman

Nov 2003

746010 Posts

Quote:
 Originally Posted by Drdmitry Given an irreducible polynomial P(x) with integer coefficients, is there any reasonable algorithm known, which constructs a random n such that P(n) is a product of two huge prime factors of relatively same size? I know how to do that for quadratic polynomials.
??? algorithm that constructs a random n ?????
If n is constructed by an algorithm, then it isn't random......

For example: P(x) = x^2 - x + 1. How does one find n such that one knows, a priori that P(n) is (say) the product of two 50-digit primes?

Last fiddled with by R.D. Silverman on 2015-09-08 at 15:49

2015-09-08, 16:17   #3
Drdmitry

Nov 2011

23410 Posts

Quote:
 Originally Posted by R.D. Silverman ??? algorithm that constructs a random n ????? If n is constructed by an algorithm, then it isn't random...... Please tell us how you do it for quadratics. For example: P(x) = x^2 - x + 1. How does one find n such that one knows, a priori that P(n) is (say) the product of two 50-digit primes?
An algorithm can take an output of a random generator as an input.
Isn't such an algorithm known for quadratics? If not then I'll write a paper on it :) . I actually did not search the literature too much for this question.

2015-09-08, 18:11   #4
R.D. Silverman

Nov 2003

11101001001002 Posts

Quote:
 Originally Posted by Drdmitry An algorithm can take an output of a random generator as an input. Isn't such an algorithm known for quadratics? If not then I'll write a paper on it :) . I actually did not search the literature too much for this question.
You claimed to KNOW how to do it for quadratics. Do you withdraw this claim?

I certainly do not know how to do it.

2015-09-08, 19:38   #5
Drdmitry

Nov 2011

3528 Posts

Quote:
 Originally Posted by R.D. Silverman 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:
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.

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

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

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

2×32×569 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.

2015-09-08, 21:34   #7
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by xilman Interesting. Please describe your algorithm in greater detail.
Indeed. This is excellent. Please tell us the method.

2015-09-09, 01:51   #8
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by R.D. Silverman Indeed. This is excellent. Please tell us the method.
One way of doing it for x^2 + 1 is to factor it as (x+i) (x-i), then search for x such that the norms
(over Q) of the factors are both prime.

For an arbitrary quadratic irreducible, factor it over its splitting field, then do as above. It may be hard to
get nearly equal primes, depending on how the poly splits.

For degree k > 2 this becomes more problematic because the polynomial now
splits into k factors.

For degree 4 use Bairstow's method to split it into quadratics [over R, with
algebraic irrational coefficients], then find the pre-image value that renders each quadratic
prime over the splitting field of the quartic. Getting the norms of each factor close may be difficult
if (say) the L2 or L_oo norms of each quadratic are quite different.

Odd degree will be difficult.

 2015-09-09, 02:44 #9 CRGreathouse     Aug 2006 134438 Posts I'd like to see more on these ideas! I haven't seen them published, and even if they exist in the folklore there's value to recording it.
2015-09-09, 03:22   #10
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by CRGreathouse I'd like to see more on these ideas! I haven't seen them published, and even if they exist in the folklore there's value to recording it.
I only gave an off-the-cuff approach using some elementary algebraic number theory.

I'd like to see the OP's ideas.

2015-09-09, 03:36   #11
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by R.D. Silverman One way of doing it for x^2 + 1 is to factor it as (x+i) (x-i), then search for x such that the norms (over Q) of the factors are both prime.
I'm an idiot. This won't work. The norm of each factor is the product of all its conjugates......
Now ask: what are the conjugates of each factor???

Still, it seems that something similar might work.... I will consider it.

 Similar Threads Thread Thread Starter Forum Replies Last Post Stargate38 Programming 18 2015-07-10 06:08 schickel FactorDB 29 2012-07-18 17:03 Dubslow Forum Feedback 0 2012-05-02 02:13 fivemack Factoring 84 2011-04-26 10:22 Xyzzy Lounge 4 2003-06-28 13:37

All times are UTC. The time now is 04:48.

Wed Sep 23 04:48:20 UTC 2020 up 13 days, 1:59, 0 users, load averages: 2.06, 2.15, 2.27

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.