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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.

Bill Dubuque
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monalisa
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  • @Mark: $9$ is not much even, is it ? –  Apr 21 '19 at 08:47
  • Yeah, didn't read the question well. Nevermind then. – Mark Apr 21 '19 at 08:48
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    It is a little weird that you needed online tools. And after just a few attempts, you should have understood why this is impossible. –  Apr 21 '19 at 08:49
  • @YvesDaoust I was just trying to cross check. That's why. – monalisa Apr 21 '19 at 08:52
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    Please edit the question to tell us what results you already know, e.g. do you know any of the named theorems in my answer? (If you don't add such context then some users may close and delete the question, so the answers will be lost). – Bill Dubuque Apr 21 '19 at 15:58

2 Answers2

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

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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 $\,\Bbb Q[x,y].$

Bill Dubuque
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