# What taking gcd of two even numbers gives

What does taking the gcd of two even numbers $y$ and $z$ give? Does it give another indefinite even number $x$?

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without further knowledge about $y$ and $z$, all we can say, is that $\gcd(y,z)$ is even. But that does not make it indefinite. – Hagen von Eitzen Oct 28 '12 at 18:50
what if both numbers are odd? – Gladstone Asder Oct 28 '12 at 18:57
If the numbers are odd then the GCD is odd, since the GCD divides both of the numbers $y$ and $z$. – Tom Oldfield Oct 28 '12 at 19:00

Any two positive integers will have a greatest common divisor which can be computed using the Euclidean algorithm, so it is a definite number. If the numbers are even, the gcd is even as well.

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Generally $\rm\ \ gcd(2j,2k)\, =\, 2\,gcd(j,k)\ \$ and $$\rm\:gcd(2j\!+\!1,2k\!+\!1)\, =\, gcd(2(j\!-\!k),2k\!+\!1)\, =\, gcd(j\!-\!k,2k\!+\!1).$$

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Imagine you have $y$ and $z$ and you make two lists: one contains all divisors of $y$ and the other contains all divisors of $z$. The greatest common divisor is just the largest number that is common to both lists.

In particular, both lists will contain the number $1$, since $1$ divides any number. Thus, $\gcd(y,z)$ is defined and is at least $1$ for any integers $y$ and $z$.

If $y$ and $z$ are both odd, it is possible that the only number appearing on both lists is $1$ (for example, $\gcd(3,5) = 1$). It is also possible for the greatest common divisor to be much larger (for example, $\gcd(10,15) = 5$).

If both $y$ and $z$ are even, the we know $2$ is a divisor of both $y$ and $z$ (this is what it means to be even), so the greatest common divisor is at least $2$. As for the odd case, however, this can be exact (e.g. $\gcd(6,8) = 2$) or very far from the truth (e.g. $gcd(32,48) = 16$).

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