I am not sure if I understand it correctly but isn't that very simple we can take gcd(2,3) = 1, and our c = 0, and gcd(0,0) does not exist?
Or my thinking process is wrong?
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That's not right. $\gcd(0,0)=0$. – Sean Eberhard Nov 23 '23 at 12:07
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Many authors leave gcd$(0, 0)$ undefined. I think it depends on the definition of gcd that is given in your course. – kabenyuk Nov 23 '23 at 15:38
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No.
Think of it this way: "$x\mid y$" is defined to be equivalent to "$y=xz$ for some $z\in D$". But $0=0z$ for all $z\in \Bbb Z$, so $0\mid 0$ (despite $\frac{0}{0}$ being undefined in this context). Divisors of $a$ are less than or equal to $a$, for all $a$. It follows that
$$\gcd(0,0)=0.$$
Shaun
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