\chapter{Dixon {\texttt{\small{[insert random buzzword here]}}} method\label{chap:dixon}} ~\cite{dixon} describes a class of ``probabilistic algorithms'' for finding a factor of any composite number, at a sub-exponential cost. They basically consists into taking random integers $r$ in $\{1, \ldots, N\}$ and look for those where $r^2 \mod{N}$ is \emph{smooth}. If enough are found, then those integers can somehow be assembled, and so a fatorization of N attemped. %% that's not really academic to be stated officially, but I would have never %% understood this section without Firas (thanks). %% %% I kept the voila` phrase, that was so lovely. \section{A little bit of History} During the latest century there has been a huge effort to approach the problem formulated by Fermat ~\ref{eq:fermat_problem} from different perspecives. This led to an entire family of algorithms known as \emph{Quadratic Sieve} [QS]. The core idea is still to find a pair of perfect squares whose difference can factorize $N$, but maybe Fermat's hypotesis can be made weaker. \paragraph{Kraitchick} was the first one popularizing the idea the instead of looking for integers $\angular{x, y}$ such that $x^2 -y^2 = N$ it is sufficient to look for \emph{multiples} of $N$: \begin{align} x^2 - y^2 \equiv 0 \pmod{N} \end{align} and, once found, claim that $\gcd(N, x \pm y)$ are non-trial divisors of $N$ just as we did in \ref{sec:fermat:implementation}. Kraitchick did not stop here: instead of trying $x^2 \equiv y^2 \pmod{N}$ he kept the value of previous attempt, and tries to find \emph{a product} of such values which is also a square. So we have a sequence \begin{align} \label{eq:dixon:x_sequence} \angular{x_0, \ldots, x_k} \mid \forall i \leq k \quad x_i^2 - N \; \text{ is a perfect square} \end{align} and hence \begin{align*} \prod_i (x_i^2 - N) = y^2 \end{align*} that $\mod{N}$ is equivalent to: \begin{align} \label{eq:dixon:fermat_revisited} y^2 \equiv \prod_i x_i^2 - N \equiv \big( \prod_i x_i \big) ^2 \pmod{N} \end{align} and voil\`a our congruence of squares. For what concerns the generation of $x_i$ with the property \ref{eq:dixon:x_sequence}, they can simply taken at random and tested using trial division. \paragraph{Brillhart and Morrison} later proposed (\cite{morrison-brillhart} p.187) a better approach than trial division to find such $x$. Their idea aims to ease the enormous effort required by the trial division. In order to achieve this. they introduce a \emph{factor base} $\factorBase$ and generate random $x$ such that $x^2 - N$ is $\factorBase$-smooth. Recalling what we anticipated in ~\ref{sec:preq:numbertheory}, $\factorBase$ is a precomputed set of primes $p_i \in \naturalPrime$. This way the complexity of generating a new $x$ is dominated by \bigO{|\factorBase|}. Now that the right side of \ref{eq:dixon:fermat_revisited} has been satisfied, we have to select a subset of those $x$ so that their product can be seen as a square. Consider an \emph{exponent vector} $v_i = (\alpha_0, \alpha_1, \ldots, \alpha_r)$ associated with each $x_i$, where \begin{align*} a_j = \begin{cases} 1 \quad \text{if $p_j$ divides $x_i$ to an odd power} \\ 0 \quad \text{otherwise} \end{cases} \end{align*} for each $0 \leq j \leq r $. There is no need to restrict ourselves for positive values of $x^2 -N$, so we are going to use $\alpha_0$ to indicate the sign. This benefit has a neglegible cost: we have to add the non-prime $-1$ to our factor base. Let now $\mathcal{M}$ be the rectangular matrix having per each $i$-th row the $v_i$ associated to $x_i$: this way each element $m_{ij}$ will be $v_i$'s $\alpha_j$. We are interested in finding set(s) of $x$ that satisfies \ref{eq:dixon:fermat_revisited}, possibly all of them. Define $K$ as the subsequence of $x_i$ whose product always have even powers. This is equivalent to look for the set of vectors $\{ w \mid wM = 0 \}$ by definition of matrix multiplication in $\mathbb{F}_2$. \paragraph{Dixon} Morrison and Brillhart's ideas of \cite{morrison-brillhart} were actually used for a slightly different factorization method, employing continued fractions instead of the square difference polynomial. Dixon refined those by porting to the quare problem, achieving a probabilistic factorization method working at a computational cost asymptotically best than all other ones previously described: \bigO{\beta(\log N \log \log N)^{\rfrac{1}{2}}} for some constant $\beta > 0$ \cite{dixon}. \section{Computing the Kernel} %%% Local Variables: %%% mode: latex %%% TeX-master: "question_authority" %%% End: