In number theory, Dixon's factorization method (also Dixon's random squares method 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.

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 ≡ y 2 ( mod N ) , x ≢ ± y ( mod N ) . {\displaystyle 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 … a n {\displaystyle a_{1}\ldots a_{n}} whose squares can be factorized as a i 2 mod N = ∏ j = 1 m b j e i j {\displaystyle a_{i}^{2}\mod N=\prod _{j=1}^{m}b_{j}^{e_{ij}}} for a fixed set b 1 … b m {\displaystyle b_{1}\ldots b_{m}} of small primes, linear algebra modulo 2 on the matrix e i j {\displaystyle e_{ij}} will give a subset of the a i {\displaystyle a_{i}} whose squares combine to a product of small primes to an even power — that is, a subset of the a i {\displaystyle 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 ai,

z 2 mod N = ∏ p i ∈ P p i a i {\displaystyle 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 ⋯ z k 2 ≡ ∏ p i ∈ P p i a i , 1 + a i , 2 + ⋯ + a i , k ( mod N ) ( where a i , 1 + a i , 2 + ⋯ + a i , k ≡ 0 ( mod 2 ) ) {\displaystyle {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

This section is taken directly from Dixon (1981).

Dixon's algorithm

Initialization. Let L be a list of integers in the range [1, n], and let P = {p1, ..., ph} be the list of the h primes ≤ v. Let B and Z be initially empty lists (Z will be indexed by B).

Step 1. If L is empty, exit (algorithm unsuccessful). Otherwise, take the first term z from L, remove it from L, and proceed to Step 2.

Step 2. Compute w as the least positive remainder of z2 mod n. Factor w as:

w = w ′ ∏ i p i a i {\displaystyle w=w'\prod _{i}p_{i}^{a_{i}}}

where w′ has no factor in P. If w′ = 1, proceed to Step 3; otherwise, return to Step 1.

Step 3. Let a ← (a1, ..., ah). Add a to B and z to Z. If B has at most h elements, return to Step 1; otherwise, proceed to Step 4.

Step 4. Find the first vector c in B that is linearly dependent (mod 2) on earlier vectors in B. Remove c from B and z c {\displaystyle z_{c}} from Z. Compute coefficients f b {\displaystyle f_{b}} such that:

c ≡ ∑ b ∈ B f b b ( mod 2 ) {\displaystyle \mathbf {c} \equiv \sum _{b\in B}f_{b}\mathbf {b} {\pmod {2}}}

Define:

d = ( d 1 , … , d n ) ← 1 2 ( c + ∑ f b b ) {\displaystyle \mathbf {d} =(d_{1},\dots ,d_{n})\gets {\frac {1}{2}}\left(\mathbf {c} +\sum f_{b}\mathbf {b} \right)}

Proceed to Step 5.

Step 5. Compute:

x ← z c ∏ b z b f b , y ← ∏ i p i d i {\displaystyle x\gets z_{c}\prod _{b}z_{b}^{f_{b}},\quad y\gets \prod _{i}p_{i}^{d_{i}}}

so that:

x 2 ≡ ∏ i p i 2 d i = y 2 mod n . {\displaystyle x^{2}\equiv \prod _{i}p_{i}^{2d_{i}}=y^{2}\mod n.}

If x ≡ y {\displaystyle x\equiv y} or x ≡ − y ( mod n ) {\displaystyle x\equiv -y{\pmod {n}}}, return to Step 1. Otherwise, return:

gcd ( n , x + y ) {\displaystyle \gcd(n,x+y)}

which provides a nontrivial factor of n, and terminate successfully.

Step-by-step example : factorizing ( n = 84923) using Dixon's algorithm

This example is lifted directly from the LeetArxiv substack. Credit is given to the original author.

  • Initialization: Define a list of numbers L, ranging from 1 to 84923:

L = { 1 , … , 84923 } {\displaystyle L=\{1,\dots ,84923\}} Define a value v, which is the smoothness factor: v = 7 {\displaystyle v=7} Define a list P containing all the prime numbers less than or equal to v: P = 2 , 3 , 5 , 7 {\displaystyle P={2,3,5,7}} Define B and Z, two empty lists. B is a list of powers, while Z is a list of accepted integers: B = [ ] {\displaystyle B=[]} Z = [ ] {\displaystyle Z=[]}

  • Step 1: Iterating z {\displaystyle z} values Write a for loop that indexes the list L {\displaystyle L}. Each element in L {\displaystyle L} is indexed as z {\displaystyle z}. The for loop exits at the end of the list.

intn=84923; for(inti=1;i<=n;i++) { intz=i; }

  • Step 2: Computing z 2 mod n {\displaystyle z^{2}\mod n} and v-smooth Prime Factorization To proceed, compute z 2 mod 84923 {\displaystyle z^{2}\mod 84923} for each z, then express the result as a prime factorization.

1 2 mod 84923 ≡ 1 mod 84923 = 2 0 ⋅ 3 0 ⋅ 5 0 ⋅ 7 0 mod 84923 {\displaystyle 1^{2}\mod 84923\equiv 1\mod 84923=2^{0}\cdot 3^{0}\cdot 5^{0}\cdot 7^{0}\mod 84923} ⋮ {\displaystyle \vdots } 513 2 mod 84923 = 8400 mod 84923 = 2 4 ⋅ 3 1 ⋅ 5 2 ⋅ 7 1 mod 84923 {\displaystyle 513^{2}\mod 84923=8400\mod 84923=2^{4}\cdot 3^{1}\cdot 5^{2}\cdot 7^{1}\mod 84923} ⋮ {\displaystyle \vdots } 537 2 mod 84923 = 33600 mod 84923 = 2 6 ⋅ 3 1 ⋅ 5 2 ⋅ 7 1 mod 84923 {\displaystyle 537^{2}\mod 84923=33600\mod 84923=2^{6}\cdot 3^{1}\cdot 5^{2}\cdot 7^{1}\mod 84923} 538 2 mod 84923 = 34675 mod 84923 = 5 2 ⋅ 19 1 ⋅ 73 1 mod 84923 {\displaystyle 538^{2}\mod 84923=34675\mod 84923=5^{2}\cdot 19^{1}\cdot 73^{1}\mod 84923}

This step continues for all values of z in the range.

  • Step 3: If z 2 mod 84923 {\displaystyle z^{2}\mod 84923} is 7-smooth, then append its powers to list B {\displaystyle B} and append z {\displaystyle z} to list Z {\displaystyle Z}.

Z = { 1 , 513 , 537 } {\displaystyle Z=\{1,513,537\}} B = { [ 0 , 0 , 0 , 0 ] , [ 4 , 1 , 2 , 1 ] , [ 6 , 1 , 2 , 1 ] } {\displaystyle B=\{[0,0,0,0],[4,1,2,1],[6,1,2,1]\}}

  • Step 4: This step is split into two parts.

Part 1: Find B {\displaystyle B} modulo 2. B = ( 0 0 0 0 4 1 2 1 6 1 2 1 ) mod 2 ≡ B = ( 0 0 0 0 0 1 0 1 0 1 0 1 ) {\displaystyle B={\begin{pmatrix}0&0&0&0\\4&1&2&1\\6&1&2&1\end{pmatrix}}\mod 2\equiv B={\begin{pmatrix}0&0&0&0\\0&1&0&1\\0&1&0&1\end{pmatrix}}}

Part 2: Check if any row combinations of B {\displaystyle B} sum to even numbers

For example, summing Row 2 {\displaystyle 2} and Row 3 {\displaystyle 3} gives us a vector of even numbers.

R 2 = { 0 , 1 , 0 , 1 } {\displaystyle R_{2}=\{0,1,0,1\}} and R 3 = { 0 , 1 , 0 , 1 } {\displaystyle R_{3}=\{0,1,0,1\}}

then R 2 + R 3 = { 0 , 1 , 0 , 1 } + { 0 , 1 , 0 , 1 } {\displaystyle R_{2}+R_{3}=\{0,1,0,1\}+\{0,1,0,1\}} R 2 + R 3 = { 0 , 2 , 0 , 2 } {\displaystyle R_{2}+R_{3}=\{0,2,0,2\}}.

  • Step 5 : This step is split into four parts. Part 1. (Finding x): Multiply the corresponding z {\displaystyle z} values for the rows found in Step 4. Then find the square root. This gives us x {\displaystyle x}. For Row 2, we had 2 4 ∗ 3 1 ∗ 5 2 ∗ 7 1 {\displaystyle 2^{4}*3^{1}*5^{2}*7^{1}}. For Row 3, we had 2 6 ∗ 3 1 ∗ 5 2 ∗ 7 1 {\displaystyle 2^{6}*3^{1}*5^{2}*7^{1}}. Thus, we find x {\displaystyle x} :

( 513 ⋅ 537 ) 2 mod 84923 = y 2 where x 2 mod 84923 = ( 513 ⋅ 537 ) 2 mod 84923 thus x = ( 513 ⋅ 537 ) mod 84923 so x = 275481 mod 84923 Finally x = 20712 mod 84923 {\displaystyle {\begin{array}{ll}(513\cdot 537)^{2}\mod 84923=y^{2}\\\\{\text{where}}\quad x^{2}\mod 84923=(513\cdot 537)^{2}\mod 84923\\\\{\text{thus}}\quad x=(513\cdot 537)\mod 84923\\\\{\text{so}}\quad x=275481\mod 84923\\\\{\text{Finally}}\quad x=20712\mod 84923\\\end{array}}}

Part 2. (Finding y): Multiply the corresponding smooth factorizations for the rows found in Step 4. Then find the square root. This gives us y {\displaystyle y}. y 2 = ( 2 4 ⋅ 3 1 ⋅ 5 2 ⋅ 7 1 ) × ( 2 6 ⋅ 3 1 ⋅ 5 2 ⋅ 7 1 ) By the multiplication law of exponents, y 2 = 2 ( 4 + 6 ) ⋅ 3 ( 1 + 1 ) ⋅ 5 ( 2 + 2 ) ⋅ 7 ( 1 + 1 ) Thus, y 2 = 2 10 ⋅ 3 2 ⋅ 5 4 ⋅ 7 2 Taking square roots on both sides gives y = 2 5 ⋅ 3 1 ⋅ 5 2 ⋅ 7 1 Therefore, y = 32 × 3 × 25 × 7 Finally, y = 16800 {\displaystyle {\begin{array}{ll}y^{2}=(2^{4}\cdot 3^{1}\cdot 5^{2}\cdot 7^{1})\times (2^{6}\cdot 3^{1}\cdot 5^{2}\cdot 7^{1})\\\\{\text{By the multiplication law of exponents,}}\\y^{2}=2^{(4+6)}\cdot 3^{(1+1)}\cdot 5^{(2+2)}\cdot 7^{(1+1)}\\\\{\text{Thus,}}\\y^{2}=2^{10}\cdot 3^{2}\cdot 5^{4}\cdot 7^{2}\\\\{\text{Taking square roots on both sides gives}}\\y=2^{5}\cdot 3^{1}\cdot 5^{2}\cdot 7^{1}\\\\{\text{Therefore,}}\\y=32\times 3\times 25\times 7\\\\{\text{Finally,}}\\y=16800\end{array}}}

Part 3. (Find x + y and x - y) where x = 20712 and y = 16800. x + y = 20712 + 16800 = 37512 {\displaystyle x+y=20712+16800=37512} x − y = 20712 − 16800 = 3912 {\displaystyle x-y=20712-16800=3912}

Part 4. Compute GCD(x+y, n) and GCD(x-y, n), where n = 84923, x+y = 292281 and x-y = 258681

gcd ( 37512 , 84923 ) = 521 gcd ( 3912 , 84923 ) = 163 {\displaystyle {\begin{array}{ll}\gcd(37512,84923)=521\\\gcd(3912,84923)=163\end{array}}}

Quick check shows 84923 = 521 × 163 {\displaystyle 84923=521\times 163}.

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 {\displaystyle \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 {\displaystyle N^{1/a}} with probability about a − a {\displaystyle 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 ⁡ ( log ⁡ N log ⁡ log ⁡ N ) {\displaystyle \exp \left({\sqrt {\log N\log \log N}}\right)}.

The optimal complexity of Dixon's method is

O ( exp ⁡ ( 2 2 log ⁡ n log ⁡ log ⁡ n ) ) {\displaystyle 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 2 ] {\displaystyle L_{n}[1/2,2{\sqrt {2}}]}

in L-notation.