
Utilisons l'algorithme des différences (et donc la propriété : $\pgcd{a}{b}=\pgcd{a}{b-a}$) pour démontrer le résultat :\vspace*{-1em}
\begin{align*}
\pgcd{x}{y} 
& = \pgcd{9k+27}{4k}\\
& \textcolor{gray}{= \pgcd{4k}{9k+27-4k}}\\
& = \pgcd{4k}{5k+27}\\
& \textcolor{gray}{= \pgcd{5k+27}{5k+27-4k}}\\
& = \pgcd{5k+27}{k+27}\\
& \textcolor{gray}{= \pgcd{k+27}{5k+27-(k+27)}}\\
& = \pgcd{k+27}{4k}\\
& \textcolor{gray}{= \pgcd{k+27}{4k-(k+27)}}\\
& = \pgcd{k+27}{3k-27}\\
& \textcolor{gray}{= \pgcd{k+27}{3k-27-(k+27)}}\\
& = \pgcd{k+27}{2k-54}\\
& \textcolor{gray}{= \pgcd{k+27}{2k-54-(k+27)}}\\
& = \pgcd{k+27}{k-81}\\
& \textcolor{gray}{= \pgcd{k+27}{k+27-(k-81)}}\\
& = \pgcd{k+27}{108}.
\end{align*}
Or, \pgcd{k+27}{108} divise 108 (par définition) donc \pgcd{x}{y} divise 108.
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