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anonymous
 4 years ago
I need ideas on how to prove that if \(a\equiv b\mod c\), then \((a,c)=(b,c)\), where \(a\), \(b\) and \(c\in\mathbb{Z}\), \(c>0\) and \((x,y)\) stands for the GCD of \(x\) and \(y\).
anonymous
 4 years ago
I need ideas on how to prove that if \(a\equiv b\mod c\), then \((a,c)=(b,c)\), where \(a\), \(b\) and \(c\in\mathbb{Z}\), \(c>0\) and \((x,y)\) stands for the GCD of \(x\) and \(y\).

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0b/a = c b/c = a if i remember correctly

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.0You should use the Euclid algorithm for finding GCD: the GCD is last nonnull remainder of the procedure. a=cq1+r, 0<= r <c b=cq2+r [the same remainder, since they are equal mod c] Now, for a the next division is c divided by r, and for b the same division ... they will lead to the same result.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I managed to prove it in three or four lines.
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