Show that \({\text{gcd}}\left( {4k + 2,\,3k + 1} \right) = {\text{gcd}}\left( {k - 1,\,2} \right)\), where \(k \in {\mathbb{Z}^ + },\,k > 1\).

State the value of \({\text{gcd}}\left( {4k + 2,\,3k + 1} \right)\) for odd positive integers \(k\).

State the value of \({\text{gcd}}\left( {4k + 2,\,3k + 1} \right)\) for even positive integers \(k\).

METHOD 1

attempting to use the Euclidean algorithm       M1

\(4k + 2 = 1\left( {3k + 1} \right) + \left( {k + 1} \right)\)      A1

\(3k + 1 = 2\left( {k + 1} \right) + \left( {k - 1} \right)\)      A1

\(K + 1 = \left( {k - 1} \right) + 2\)      A1

\( = {\text{gcd}}\left( {k - 1,\,2} \right)\)     AG

METHOD 2

\({\text{gcd}}\left( {4k + 2,\,3k + 1} \right)\)

\( = {\text{gcd}}\left( {4k + 2 - \left( {3k + 1} \right),\,3k + 1} \right)\)     M1

\( = {\text{gcd}}\left( {3k + 1,\,k + 1} \right)\,\,\left( { = {\text{gcd}}\left( {{\text{k + 1,}}\,{\text{3k + 1}}} \right)} \right)\)     A1

\( = {\text{gcd}}\left( {3k + 1 - 2\left( {k + 1} \right),\,k + 1} \right)\,\,\left( { = {\text{gcd}}\left( {k - 1{\text{,}}\,k + {\text{1}}} \right)} \right)\)     A1

\( = {\text{gcd}}\left( {k + 1 - \left( {k - 1} \right),\,k - 1} \right)\,\,\left( { = {\text{gcd}}\left( {{\text{2,}}\,k - {\text{1}}} \right)} \right)\)     A1

\( = {\text{gcd}}\left( {k - 1,\,2} \right)\)     AG

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(for \(k\) odd), \({\text{gcd}}\left( {4k + 2,\,3k + 1} \right) = 2\)     A1

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(for \(k\) even), \({\text{gcd}}\left( {4k + 2,\,3k + 1} \right) = 1\)     A1

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