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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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