Dixon's factorization method

In number theory, Dixon's factorization method (also Dixon's random squares method^[1] or Dixon's algorithm) is a general-purpose integer factorization algorithm; it is the prototypical factor base method. Unlike for other factor base methods, its run-time bound comes with a rigorous proof that does not rely on conjectures about the smoothness properties of the values taken by a polynomial.

The algorithm was designed by John D. Dixon, a mathematician at Carleton University, and was published in 1981.^[2]

Basic idea

Dixon's method is based on finding a congruence of squares modulo the integer N which is intended to factor. Fermat's factorization method finds such a congruence by selecting random or pseudo-random x values and hoping that the integer x² mod N is a perfect square (in the integers):

x^{2}\equiv y^{2}\quad ({\hbox{mod }}N),\qquad x\not \equiv \pm y\quad ({\hbox{mod }}N).

For example, if N = 84923, (by starting at 292, the first number greater than √N and counting up) the 505² mod 84923 is 256, the square of 16. So (505 − 16)(505 + 16) = 0 mod 84923. Computing the greatest common divisor of 505 − 16 and N using Euclid's algorithm gives 163, which is a factor of N.

In practice, selecting random x values will take an impractically long time to find a congruence of squares, since there are only √N squares less than N.

Dixon's method replaces the condition "is the square of an integer" with the much weaker one "has only small prime factors"; for example, there are 292 squares smaller than 84923; 662 numbers smaller than 84923 whose prime factors are only 2,3,5 or 7; and 4767 whose prime factors are all less than 30. (Such numbers are called B-smooth with respect to some bound B.)

If there are many numbers $a_{1}\ldots a_{n}$ whose squares can be factorized as $a_{i}^{2}\mod N=\prod _{j=1}^{m}b_{j}^{e_{ij}}$ for a fixed set $b_{1}\ldots b_{m}$ of small primes, linear algebra modulo 2 on the matrix $e_{ij}$ will give a subset of the $a_{i}$ whose squares combine to a product of small primes to an even power — that is, a subset of the $a_{i}$ whose squares multiply to the square of a (hopefully different) number mod N.

Method

Suppose the composite number N is being factored. Bound B is chosen, and the factor base is identified (which is called P), the set of all primes less than or equal to B. Next, positive integers z are sought such that z² mod N is B-smooth. Therefore we can write, for suitable exponents a_i,

z^{2}{\text{ mod }}N=\prod _{p_{i}\in P}p_{i}^{a_{i}}

When enough of these relations have been generated (it is generally sufficient that the number of relations be a few more than the size of P), the methods of linear algebra, such as Gaussian elimination, can be used to multiply together these various relations in such a way that the exponents of the primes on the right-hand side are all even:

{z_{1}^{2}z_{2}^{2}\cdots z_{k}^{2}\equiv \prod _{p_{i}\in P}p_{i}^{a_{i,1}+a_{i,2}+\cdots +a_{i,k}}\ {\pmod {N}}\quad ({\text{where }}a_{i,1}+a_{i,2}+\cdots +a_{i,k}\equiv 0{\pmod {2}})}

This yields a congruence of squares of the form a² ≡ b² (mod N), which can be turned into a factorization of N, N = gcd(a + b, N) × (N/gcd(a + b, N)). This factorization might turn out to be trivial (i.e. N = N × 1), which can only happen if a ≡ ±b (mod N), in which case another try must be made with a different combination of relations; but if a nontrivial pair of factors of N is reached, the algorithm terminates.

Pseudocode

input: positive integer  ${\textstyle N}$ 
output: non-trivial factor of  ${\textstyle N}$ 

Choose bound  ${\textstyle B}$ 
Let  ${\textstyle P:=\{p_{1},p_{2},\ldots ,p_{k}\}}$  be all primes  ${\textstyle \leq B}$ 

repeat
    for  ${\textstyle i=1}$  to  ${\textstyle k+1}$  do
        Choose  ${\textstyle 0<z_{i}<N}$  such that  ${\textstyle z_{i}^{2}{\text{ mod }}N}$  is  ${\textstyle B}$ -smooth
        Let  ${\textstyle a_{i}:=\{a_{i1},a_{i2},\ldots ,a_{ik}\}}$  such that  ${\textstyle z_{i}^{2}{\text{ mod }}N=\prod _{p_{j}\in P}p_{j}^{a_{ij}}}$ 
    end for

    Find non-empty  ${\textstyle T\subseteq \{1,2,\ldots ,k+1\}}$  such that  ${\textstyle \sum _{i\in T}a_{i}\equiv {\vec {0}}{\pmod {2}}}$ 
    Let  ${\textstyle x:=\left(\prod _{i\in T}z_{i}\right){\text{ mod }}N}$ 
         ${\textstyle y:=\left(\prod _{p_{j}\in P}p_{j}^{\left(\sum _{i\in T}a_{ij}\right)/2}\right){\text{ mod }}N}$ 
while  ${\textstyle x\equiv \pm y{\pmod {N}}}$  

return  ${\textstyle \gcd(x+y,N)}$

Example

This example will try to factor N = 84923 using bound B = 7. The factor base is then P = {2, 3, 5, 7}. A search can be made for integers between $\left\lceil {\sqrt {84923}}\right\rceil =292$ and N whose squares mod N are B-smooth. Suppose that two of the numbers found are 513 and 537:

513^{2}\mod 84923=8400=2^{4}\cdot 3\cdot 5^{2}\cdot 7

537^{2}\mod 84923=33600=2^{6}\cdot 3\cdot 5^{2}\cdot 7

So

(513\cdot 537)^{2}\mod 84923=2^{10}\cdot 3^{2}\cdot 5^{4}\cdot 7^{2}\mod 84923

Then

${\begin{aligned}&{}(513\cdot 537)^{2}\mod 84923\\&=(275481)^{2}\mod 84923\\&=(84923\cdot 3+20712)^{2}\mod 84923\\&=(84923\cdot 3)^{2}+2\cdot (84923\cdot 3\cdot 20712)+20712^{2}\mod 84923\\&=0+0+20712^{2}\mod 84923\end{aligned}}$

That is, $20712^{2}\mod 84923=(2^{5}\cdot 3\cdot 5^{2}\cdot 7)^{2}\mod 84923=16800^{2}\mod 84923.$

The resulting factorization is 84923 = gcd(20712 − 16800, 84923) × gcd(20712 + 16800, 84923) = 163 × 521.

Optimizations

The quadratic sieve is an optimization of Dixon's method. It selects values of x close to the square root of N such that x² modulo N is small, thereby largely increasing the chance of obtaining a smooth number.

Other ways to optimize Dixon's method include using a better algorithm to solve the matrix equation, taking advantage of the sparsity of the matrix: a number z cannot have more than $\log _{2}z$ factors, so each row of the matrix is almost all zeros. In practice, the block Lanczos algorithm is often used. Also, the size of the factor base must be chosen carefully: if it is too small, it will be difficult to find numbers that factorize completely over it, and if it is too large, more relations will have to be collected.

A more sophisticated analysis, using the approximation that a number has all its prime factors less than $N^{1/a}$ with probability about $a^{-a}$ (an approximation to the Dickman–de Bruijn function), indicates that choosing too small a factor base is much worse than too large, and that the ideal factor base size is some power of $\exp \left({\sqrt {\log N\log \log N}}\right)$ .