## Algorithm

To factor a number $n$ using the quadratic sieve  , one seeks two numbers $x$ and $y$ which are not congruent modulo $n$ with $x$ not congruent  to $-y$ modulo $n$ but have $x^{2}\equiv y^{2}\pmod{n}$. If two such numbers are found, one can then say that $(x+y)(x-y)\equiv 0\pmod{n}$. Then, $x+y$ and $x-y$ must have non-trivial factors in common with $n$.

To accomplish this, the quadratic sieve method uses a set of prime numbers  called a factor base. Then, it searches for numbers which can be factored entirely within that factor base. If there are $k$ prime numbers in the factor base, then each number which can be factored within the factor base is stored as a $k$-dimensional vector where the $i$-th component    of the vector for $y$ gives the exponent of the $i$-th prime from the factor base in the factorization of $y$. For example, if the factor base were $\left\{2,3,5,7,11,13\right\}$, then the number $y=2^{3}\cdot 3^{2}\cdot 11^{5}$ would be stored as the vector $\left<3,2,0,0,5,0\right>$.

Once $k+1$ of these vectors have been collected, there must be a linear dependence among them. The $k+1$ vectors are taken modulo $2$ to form vectors in $\mathbb{Z}_{2}^{k}$. The linear dependence among them is used to find a combination   of the vectors which sum up to the zero vector  in $\mathbb{Z}_{2}^{k}$. Summing these vectors is equivalent to multiplying the $y$’s to which they correspond. And, the zero vector in $\mathbb{Z}_{2}^{k}$ signals a perfect square  .

To factor $n$, chose a factor base $\mathbf{B}=\left\{p_{1},p_{2},\ldots,p_{k}\right\}$ such that $2\in\mathbf{B}$ and for each odd prime $p_{j}$ in $\mathbf{B}$, $n$ is a quadratic residue  of $p_{j}$. Now, start picking $x_{i}$ near $\sqrt{n}$ and calculate $y_{i}=x_{i}^{2}-n$. Clearly $y_{i}\equiv x_{i}^{2}\pmod{n}$. If $y_{i}$ can be completely factored by numbers in $\mathbf{B}$, then it is called $\mathbf{B}$-smooth. If it is not $\mathbf{B}$-smooth, then discard $x_{i}$ and $y_{i}$ and move on to a new choice of $x_{i}$. If it is $\mathbf{B}$-smooth, then store $x_{i}$, $y_{i}$, and the vector of its exponents for the primes in $\mathbf{B}$. Also, record a copy of the exponent vector with each component taken modulo $2$.

Once $k+1$ vectors have been recorded, there must be a linear dependence among them. Using the copies of the exponent vectors that were taken modulo $2$, determine which ones can be added together to form the zero vector. Multiply together the $x_{i}$ that correspond to those chosen vectors—call this $x$. Also, add together the original vectors that correspond to the chosen vectors to form a new vector $\vec{v}$. Every component of this vector will be even. Divide each element of $\vec{v}$ by $2$. Form $y=\prod_{i=1}^{k}p_{i}^{v_{i}}$.

Because each $y_{i}\equiv x_{i}^{2}\pmod{n}$, $x^{2}\equiv y^{2}\pmod{n}$. If $x\equiv y\pmod{n}$, then find some more $\mathbf{B}$-smooth numbers and try again. If $x$ is not congruent to $y$ modulo $n$, then $(x+y)$ and $(x-y)$ are factors of $n$.

## Example

Consider the number $n=16843009$ The integer nearest its square root is $4104$. Given the factor base

 $\mathbf{B}=\left\{2,3,5,7,13\right\}$

, the first few $\mathbf{B}$-smooth values of $y_{i}=f(x_{i})=x_{i}^{2}-n$ are:

$x_{i}$ $y_{i}=f(x_{i})$ 2 3 5 7 13
4122 147875 0 0 3 1 2
4159 454272 7 1 0 1 2
4187 687960 3 3 1 2 1
4241 1143072 5 6 0 2 0
4497 3380000 5 0 4 0 2
4993 8087040 9 5 1 0 1

Using $x_{0}=4241$ and $x_{1}=4497$, one obtains:

 $y_{0}=1143072=2^{5}\cdot 3^{6}\cdot 5^{0}\cdot 7^{2}\cdot 13^{0}$
 $y_{1}=3380000=2^{5}\cdot 3^{0}\cdot 5^{4}\cdot 7^{0}\cdot 13^{2}$

Which results in:

 $x=4241\cdot 4497=19071777$
 $y=2^{5}\cdot 3^{3}\cdot 5^{2}\cdot 7^{1}\cdot 13^{1}=1965600$

From there:

 $\gcd(x-y,n)=257$
 $\gcd(x+y,n)=65537$

It may not be completely obvious why we required that $n$ be a quadratic residue of each $p_{i}$ in the factor base $\mathbf{B}$. One might intuitively think that we actually want the $p_{i}$ to be quadratic residues of $n$ instead. But, that is not the case.

We are trying to express $n$ as:

 $(x+y)(x-y)=x^{2}-y^{2}=n$

where

 $y=\prod_{i=1}^{k}p_{i}^{v_{i}}$

Because we end up squaring $y$, there is no reason that the $p_{i}$ would need to be quadratic residues of $n$.

So, why do we require that $n$ be a quadratic residue of each $p_{i}$? We can rewrite the $x^{2}-y^{2}=n$ as:

 $x^{2}-\prod_{i=1}^{k}p_{i}^{2v_{i}}=n$

If we take that expression modulo $p_{i}$ for any $p_{i}$ for which the corresponding $v_{i}$ is non-zero, we are left with:

 $x^{2}\equiv n\pmod{p_{i}}$

Thus, in order for $p_{i}$ to show up in a useful solution, $n$ must be a quadratic residue of $p_{i}$. We would be wasting time and space to employ other primes in our factoring and linear combinations  .

Title quadratic sieve QuadraticSieve 2013-03-22 12:48:14 2013-03-22 12:48:14 patrickwonders (217) patrickwonders (217) 9 patrickwonders (217) Algorithm  msc 11Y05 msc 11A51