Normal Euclidean algorithm iterates $(a$ mod $b, b)$ for $a>b$. I have noticed that iterating $(a$ mod $ b, a)$ (keeping the $a$ fixed through descent) will also give the GCD. Can anyone show why is that so? I have no experience in proving.
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This is Gauss's algorithm and generally it works only for prime $a.\ $ Did you forget the many earlier discussions on this in your older posts circa $5$ months ago? – Bill Dubuque Jul 04 '19 at 20:13
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How does this work when $a = 9$ and $b = 6$? – Eric Towers Jul 04 '19 at 20:14
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I went back to this after my post from 2 days ago. Actually I was trying to understand why when $p|ab$ the $p|(p$ mod $ a)b$ – Michael Munta Jul 04 '19 at 20:19
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Hint $,\ p\mid pb,ab,\Rightarrow, p\mid (\color{#c00}{p!-!ka})b.\ $ But $,\color{#c00}{p!-!ka},=, p\bmod a, $ for some $k\ \ $ – Bill Dubuque Jul 04 '19 at 20:45
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@BillDubuque is it possible I can contact you via email or some other portal where you are active which allows private messages? – Michael Munta Jul 05 '19 at 07:42
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I wish to ask you about your history in math, which college you went to, how much did you self study etc. I like how you always have a different view on stuff. – Michael Munta Jul 05 '19 at 09:26
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@Michael Remove the at-sign and append google.com to my username to get my mail address – Bill Dubuque Jul 05 '19 at 12:10
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@Bill can you please confirm if you received my email? – Michael Munta Jul 05 '19 at 17:28
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@EricTowers well it would work for $a = 9$ and $b = 6$, but it would not for example for $a = 14$ and $b = 5$. As Bill said it will always work for prime $a$. – Michael Munta Jul 11 '19 at 14:51