Integer factorization is the process of decomposing a composite number into a product of smaller integers, if these integers are restricted to be prime numbers then the process is called prime factorization

The fundamental theorem of arithmetic states that ever positive integer can be written uniquely as a product of primes

example:

$$65340 = 2^2 \cdot 3^3 \cdot 5 \cdot 11^2$$

## Trial division

Trial division is the simplest algorithm for factoring an integer, we assume that $s$ and $t$ are factors of a number $n$ such that $n = st$ and $s \leq t$ (note that $s$ and $t$ do not need to be prime numbers), when a divisor $s$ is found then $n / s$ is also a factor

vector<int> trial_division(int n) {
for (int i = 2; i * i <= n; i += 1) {
if (n % i == 0) {
// n is a composite number
return vector<int> {i, n / i};
}
}
// n is a prime number
return vector<int> {n};
}


## Fermat factorization

Fermat’s observation was to write an integer as the difference of squares

\begin{align} n &= x^2 - y^2 \label{fermat} \\ &= (x + y)(x - y) \end{align}

Assuming that $s$ and $t$ are odd factors of $n$ such that $n = st$ and $s \leq t$ we can find $x$ and $y$ such that

\begin{align*} s &= x - y \\ t &= x + y \end{align*}

$$s + t = 2x \\ x = \frac{s + t}{2}$$

Also

$$y = \frac{t - s}{2}$$

Since we assumed that $s$ and $t$ are odd numbers, their difference is an even number which is divisible by $2$ therefore $x$ and $y$ are integers, since $s > 1$ and $t \geq s$ we find that $x \geq 1$ and $y \geq 0$

From \eqref{fermat} we also know that $x = \sqrt{n + y^2}$ and hence $x \geq \sqrt{n}$, also $x = \tfrac{s + t}{2}$ and we know that the upper bound of $s$ happens when $s$ is as close as $t$ as possible, given that $s \leq t$, $x \leq \tfrac{t + t}{2} \leq n$

Implementation notes: since $s$ and $t$ are odd numbers, their product $n$ is also an odd number, therefore the implementation below works with odd values of $n$

/**
* Factorization of an odd number n based on Fermat's
* factorization algorithm
*
* @param  {int} n
* @return {vector<int>} a vector with two odd integers if n is not a
* prime number, a single integer if n is a prime number
*/
vector<int> fermat_factorization(int n) {
for (int x = (int) ceil(sqrt(n)); x <= n; x += 1) {
int ySquared = x * x - n;
// check if y is the square of some number
int y = (int) sqrt(ySquared);
if (y * y == ySquared) {
int s = x - y;
int t = x + y;
// s must be > 1
if (s != 1 && t != n) {
return vector<int> {s, t};
}
}
}
// n is a prime number
return vector<int> {n};
}


### Pollard’s Rho factorization

Pollard’s Rho factorization is a probabilistic factorization algorithm based on the assumption that a number $n$ is a composite number and the following facts:

• since $n$ is a composite number there must be a factor $d$
• let $a$, $b$ two positive integers, if $a \equiv b \pmod{d}$ then the difference $a - b$ is a multiple of $d$, since $n$ is also a multiple of $d$ some multiple of $d$ is a divisor of $n$ and $a - b$, particularly $gcd(a - b, n)$ is a divisor of $n$, let $gcd(a - b, n) > 1$ then we have found two factors of $n$ ($gcd(a - b, n), \tfrac{n}{gcd(a - b, n)}$)

Now the problem is reduced to find $a$ and $b$ such that $gcd(a - b, n) > 1$, we can use the following algorithm which picks random numbers in the range $[1, n - 1]$

let n be the number to be factorized
let x be an array of integers

x = random integer in the range [1, n - 1]
while we haven't two numbers such that gcd(x_i, x_j, n) > 1
x[i] = random integer in the range [1, n - 1]
for all j < i and j >= 0
if gcd(x[i] - x[j], n) > 1
return x[i], x[j]


This page has a good explanation on how this algorithm will find $a$ and $b$ with a probability $~50%$ after $\sqrt{n}$ iterations, the algorithm above is not very helpful though since at the $k$ iteration we have to do $k - 1$ pairwise checks

Here’s another algorithm to pick random numbers, let $x$ be an integer in the range $[1, n - 1]$, a function that will generate a number in the range $[1, n - 1]$ based on a previous number is

$$f(x) = x^2 + c \pmod{n}$$

Because there are only $n - 1$ possible values our generator will eventually fall into a cycle, for example let $n = 55, c = 2, x = 2$

\begin{align*} x_0 &= 2 \\ x_1 &= (2^2 + 2) \pmod{55} = 6 \\ x_2 &= (6^2 + 2) \pmod{55} = 38 \\ x_3 &= (38^2 + 2) \pmod{55} = 16 \\ x_4 &= (16^2 + 2) \pmod{55} = 38 \text{ which is equal to x_2 } \end{align*}

Pollard detected the cycle using Floyd’s cycle-finding algorithm which is based on two pointers which move through a sequence at different speeds, one moves a unit and the other moves two units each time, if there’s a cycle eventually the two pointers will encounter at some element belonging to the cycle, if we’ve analyzed all the elements of the sequence and saw not a single contiguous pair fullfills $gcd(x_i - x_{i + 1}, n) > 1$ we need to choose other values for $x_0, a$ and rerun the algorithm

// C++11
#include <random>

/**
* Computes a factor of n which is greater than 1
* @param {long long} n The number to be factorized
* @return {long long} A positive integer which is a factor of n
* when the algorithm successfully finds a factor of n, -1 otherwise
*/
long long pollard_rho(long long n) {
if (n % 2 == 0) {
return 2;
}

std::random_device rd;
std::mt19937 engine(rd());
std::uniform_int_distribution<long long> dis(1, n - 1);

long long x = dis(engine);
long long c = dis(engine);
long long y = x;   // y = x^2 + c (mod n)
do {
// tortoise goes x = f(x)
x = ((x * x) % n + c) % n;

// hare goes y = f(f(y))
y = ((y * y) % n + c) % n;
y = ((y * y) % n + c) % n;

long long gcd = __gcd(abs(x - y), n);
if (gcd > 1) {
return gcd;
}
} while (x != y);

return -1;
}

/**
* Pollard rho factorization runner, it makes multiple calls to
* pollard_rho until a factor is found
* @param {long long} n The number to be factorized
* @return {long long} A factor of n (it is n when n is a prime number)
*/
long long pollard_rho_factorization(long long n) {
long long factor;
do {
factor = pollard_rho(n);
} while(factor < 0);
return factor;
}


### Eratosthenes Sieve factorization of a range

We can also compute the factorization of a number by modifying the sieve of Erathostenes, remember that each state of the sieve hold a boolean telling if the number is prime or not, this time each state of the sieve will hold a pair of numbers

• the lowest prime that is a divisor of any index i
• the maximum power of the lowest prime computed above (optional)

Let’s represent $n$ as $p_1^{a_1} \cdot p_2^{a_2} \ldots p_n^{a_n}$, since we’re hold for each position the lowest prime and its the maximum power, the state stored at the position $n$ of the sieve will be $p_1^{a_1}$, if we divide $n$ by this number we will move to the state $p_2^{a_2} \ldots p_n^{a_n}$, this recursive process is run until the current state reached in the sieve is $1$

// let p be the smallest prime factor of the index i, each element
// contains a pair (p, x) such that p^x is a divisor of i
// e.g.
//
//    8 = (2, 3)
//    15 = (3, 1)
//    6 = (2, 1)
//
vector<pair<int, int> > lp;

void eratosthenes_sieve_factorization(long long n) {
pair<int, int> unvisited(-1, -1);

// (-1, -1) is an unvisited state
lp.assign(n + 1, unvisited);

for (int i = 2; i * i <= n; i += 1) {
if (lp[i] == unvisited) {
// if an index is in an unvisited state it's a prime number
pair<int, int> base(i, 1);
lp[i] = base;
for (int j = i * i; j <= n; j += i) {
if (lp[j] == unvisited) {
// if a multiple of the prime number is in an unvisited
// state that means that it's lowest prime divisor is
// the current index i
lp[j] = base;
if (lp[j / i] != unvisited) {
// j is a multiple of i, in fact j = i^x because
// only numbers which don't have other factors but i
// reach this point, to accumulate the powers it's enough
// find out the power of j / i
lp[j].second += lp[j / i].second;
}
}
}
}
}

// all the prime numbers > sqrt(n) will have an unvisited state
// changing the unvisited state to prime_number^1
int sqrtN = sqrt(n);
if (sqrtN % 2 == 0) {
sqrtN += 1;
}
for (int i = sqrtN; i <= n; i += 2) {
if (lp[i] == unvisited) {
lp[i] = pair<int, int>(i, 1);
}
}
}