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Let $x,y,z \in \mathbb{Z}$ where $x \neq 0$ or $y \neq 0$. If $z$ is a common divisor and $z \neq 0$, prove $$ gcd(x,y) / |z| = gcd(x/z, y/z)$$

I realize this can be written as $$ gcd( x/|z|, y / |z|) = gcd(x/z, y/z) $$ But I am unsure as to how to continue as $ |z| \neq z$ .

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    $ax+by=d \iff ax/|z|+by/|z|=d/|z|$ – 2017-02-14

1 Answers 1

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To see this, use Bezout's theorem. Let $g = \gcd(x,y)$. Then, there exist integers $a,b$ such that $g = ax+by$.

Since $|z|$ divides $g$, we have that $g/|z| = a(x/|z|) + b(y/|z|)$.

The above shows that $\gcd(x/|z|, y/|z|) \leq g/|z|$.

However, the other way is also true: suppose that $h = \gcd(x/|z|, y/|z|)$, then there exist integers $a,b$ such that $h = a(x/|z|) + b(y/|z|)$. Multiplying by $|z|$, we get that $h|z| = ax + by$. Therefore, we see that $h|z| \geq g$, by Bezout's theorem.

Combining these, $h=g$.