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@@ -44,9 +44,9 @@ that $\mod{N}$ is equivalent to:
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\label{eq:dixon:fermat_revisited}
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y^2 \equiv \prod_i (x_i^2 - N) \equiv \big( \prod_i x_i \big) ^2 \pmod{N}
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\end{align}
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-and voil\`a our congruence of squares. For what concerns the generation of $x_i$
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-with the property \ref{eq:dixon:x_sequence}, they can simply taken at random and
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-tested using trial division.
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+and voil\`a our congruence of squares (\cite{discretelogs} \S 4). For what
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+concerns the generation of $x_i$ with the property \ref{eq:dixon:x_sequence},
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+they can simply taken at random and tested using trial division.
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\paragraph{Brillhart and Morrison} later proposed (\cite{morrison-brillhart}
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p.187) a better approach than trial division to find such $x$. Their idea aims
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@@ -225,10 +225,10 @@ $e^{\sqrt{\ln N \ln \ln N}}$.
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\State $x_i \getsRandom \{0, \ldots N\}$
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\State $y_i \gets x_i^2 - N$
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\State $v_i \gets \texttt{smooth}(y_i)$
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- \If{$v_i \neq \emptyset$} $i++$ \EndIf
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+ \If{$v_i$} $i \gets i+1$ \EndIf
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\EndWhile
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\State $\mathcal{M} \gets \texttt{matrix}(v_0, \ldots, v_f)$
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- \For{$\angular{\lambda_0, \ldots, \lambda_k}
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+ \For{$\lambda = \{\mu_0, \ldots, \mu_k\}
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\strong{ in } \texttt{ker}(\mathcal{M})$}
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\Comment get relations
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\State $x \gets \prod\limits_{\mu \in \lambda} x_\mu \pmod{N}$
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