An example of an even numbers $n,k$ whose $gcd$ is three.

Question

I am looking for even integers $n$ and $k$ such that $k$ does not divide $n$ and $\gcd(n,k) = 3$. Is this possible?

With the help of some online tools I tried, but every time I am not getting the desired result. Kindly help.

Answer 1

Even numbers are all multiples of $2$, hence the $\text{gcd}$ is perforce even.

Answer 2

The greatest common divisor of even integers is even. Though this follows easily from deeper GCD properties it also has a simple direct proof. Suppose $c$ is an odd common divisor of $2a$ and $2b$. By the Lemma below we infer that $2c$ is also a common divisor, so $c$ is not the greatest common divisor.

Lemma $\ $ Suppose that an odd integer $\,c\,$ divides $\,2k$. Then if follows that $\,2c$ divides $2k$

Proof $\ $ By hypothesis $\,nc = 2k.\,$ So $\,nc\,$ even, $c$ odd $\,\Rightarrow\, n$ even, so $\,(n/2)\, 2c = 2k$

Remark $ $ Alternatively it follows immediately from basic GCD properties (assuming they are known).

By the GCD Universal Property $\ \ d\mid n,k\iff d\mid\gcd(n,k),\ $ Put $\, d = 2.\ \ $

By the GCD Distributive Law $\ \gcd(2a,2b) = 2\gcd(a,b)\,$ is even.

By the GCD Bezout Identity $\ \ \ \gcd(2a,2b) = j(2a)+k(2b) = 2(ja+kb)$

The Bezout-based proof is less general than the prior two proofs since it need not hold in domains where gcds exist, e.g. in polynomial rings such as $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y].$