Talk:Dixon's factorization method

Comment
I cannot see a real difference to the Quadratic Sieve, but thar does not really matter.
 * Dixon's method is not a well-specified algorithm in that it doesn't really give a method for finding relations (presumably, brute-force search would be used). The quadratic sieve uses some mathematical facts and the sieving procedure to speed up the finding of relations. The number field sieve is yet another way of finding relations. Decrypt3 20:47, 24 March 2006 (UTC)
 * The confusion between Dixon's method and quadratic sieve is something I encounter increasingly frequently in the wild. I once saw a video on YouTube where a guy explains Dixon's method in great detail, all the while believing he is explaining quadratic sieve. Then just today I found another video on the same topic by the same guy, with the same mistake. And it's in this article too, where it asserts that the strength of QS is in finding small values of k^2 mod N, while its actual strength is in avoiding separate factorization of each candidate, and in spending very little time on dealing with failing candidates. I plan to fix this tomorrow. Rulatir (talk) 00:01, 21 October 2021 (UTC)

Question
I didn't understand this sentence: "This set of primes is called the factor base. Then, using the polynomial p(x) = x2 − n, many values of x are tested to see if p(x) factors completely over the factor base."

What is x? What is n?  What does it mean to factor "over the factor base"?

n is the semi-prime which we are trying to factor x is one of many random numbers generated to try and find a p(x) which the prime factorization is contained in the factor base. for example if x=20 The prime factorization of 20 is: 2*2*5 If our factor base is [2,3,5] then p(x) factors completley or is "smooth" over our factor base. I'm curious though is p(x)=x^2-n suposed to be p(x)=x^2 (mod n)

Overlap
This article has quite a bit of overlap with quadratic sieve, which also goes into quite a bit of detail regarding the linear algebra stage. This should be either in one or the other, or moved out to congruence of squares. Dcoetzee 02:47, 19 May 2007 (UTC)

Example is slightly faked
When testing N=84923 one actully find first b-smooth z=436, 436^2 mod N = 20250, which is 2*3^4*5^3.

Another important fact is that, we find in this example z=505 (less than "first" 513), is square 505^2 mod N = 256, which is 2^8. So whole procedure is somehow useless.

Maybe slightly better number N could be find to better ilustrate its application of Dixon's method.

--149.156.82.207 (talk) 20:36, 8 December 2010 (UTC)

Reference for running time?
What is the reference for the running time of L[1/2, 2sqrt(2)] claimed in the article? In Dixon's paper the time is stated as L[1/2, 3sqrt(2)], while Pomerance arrives at L[1/2, 2] for the plain method in (bound is claimed to be sharp there). --2003:51:4F60:6F01:D056:A97A:7C5F:91ED (talk) 19:38, 11 December 2014 (UTC)