# Dixon's factorization method

(Redirected from Dixon's algorithm)

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, and the only factor base method for which a run-time bound not reliant on conjectures about the smoothness properties of values of a polynomial is known.

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 we intend to factor. Fermat's factorization algorithm 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, we notice (by starting at 292, the first number greater than N and counting up) that 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 us 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 we have lots of 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 us 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 we are trying to factor the composite number N. We choose a bound B, and identify the factor base (which we will call P), the set of all primes less than or equal to B. Next, we search for positive integers z such that z2 mod N is B-smooth. We can therefore write, for suitable exponents ak,

$z^2 \equiv \prod_{p_i\in P} p_i^{a_i} \pmod{N}$

When we have generated enough of these relations (it's generally sufficient that the number of relations be a few more than the size of P), we can use the methods of linear algebra (for example, Gaussian elimination) 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 gives us a congruence of squares of the form a2b2 (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 we have to try again with a different combination of relations; but with luck we will get a nontrivial pair of factors of N, and the algorithm will terminate.

## Example

We will try to factor N = 84923 using bound B = 7. Our factor base is then P = {2, 3, 5, 7}. We then search randomly for integers between $\left\lceil\sqrt{84923} \right\rceil = 292$ and N whose squares are B-smooth. Suppose that two of the numbers we find 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

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

Then

\begin{align} & {} \qquad (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{align}

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 $\exp\left(\sqrt{\log N \log \log N}\right)$.

The optimal complexity of Dixon's method is

$O\left(\exp\left(2 \sqrt 2 \sqrt{\log n \log \log n}\right)\right)$

in big-O notation, or

$L_n [1/2, 2 \sqrt 2]$

in L-notation.

## References

1. ^ Kleinjung, Thorsten; et al. (2010). "Factorization of a 768-bit RSA modulus". Advances in Cryptology – CRYPTO 2010. Lecture Notes in Computer Science 6223. pp. 333–350. doi:10.1007/978-3-642-14623-7_18.
2. ^ Dixon, J. D. (1981). "Asymptotically fast factorization of integers". Math. Comp. 36 (153): 255–260. doi:10.1090/S0025-5718-1981-0595059-1. JSTOR 2007743.