# Show that if $\gcd(abc,d^2)=1$, then $\gcd(a,d)=\gcd(b,d)=\gcd(c,d)=1$.

Let $$a,b,c$$ be integers. Show that if $$\gcd(abc,d^2)=1$$, then $$\gcd(a,d)=\gcd(b,d)=\gcd(c,d)=1$$.

Here is my way of approaching this question:

Suppose $$\gcd(abc,d^2)=1$$, there exist integers $$x,y$$ such that $$abcx+d^2y=1$$

$$a(bcx)+d(dy)=1$$, which implies that $$\gcd(a,d)=1$$

$$b(acx)+d(dy)=1$$, which implies that $$\gcd(b,d)=1$$

$$c(abx)+d(dy)=1$$, which implies that $$\gcd(c,d)=1$$

So far I don't really know if this is the way to answer this question. Any help would be appreciated.

• your proof is perfect ! – Surb Feb 16 at 18:16

Yes this is a valid use of Bézout's identity to solve the problem.

Yes, that's a correct Bezout-based proof. More generally we can used basic gcd laws such as below to deduce that $$\ (a,d)\mid (abc,d^2)=1\$$ so $$\,(a,d)=1,\$$ with $$\,(x,y) := \gcd(x,y)\,\$$ [standard notation]

Lemma $$\ a\mid A,\ b\mid B\,\Rightarrow (a,b)\mid (A,B)\$$

Proof $$\ \ \ (a,b)\mid A,B\,\Rightarrow\, (a,b)\mid (A,B)$$

Remark  We could also prove the lemma using the Bezout identity for $$(A,B) = 1\,$$ just as you did. This yields a unified proof for all three inferences. But the Bezout-based proofs fail in more general rings where gcds exist but they are not of Bezout linear form, e.g. polynomial rings $$\Bbb Z[x]$$ or $$\,\Bbb Q[x,y],\,$$ where $$\,(x,y) = 1\,$$ but this gcd has no Bezout linear representation $$\,xg(x,y) + y f(x,y) = 1,\,$$ else evaluation at $$\,x = y = 0\,$$ yields $$\,0 = 1.\,$$

Yes, your proof is correct. You can also do like this:

Say $$\gcd(a,d)=r$$

Then $$d =rd'$$ and $$a=ra'$$

Then $$abc = ra'bc$$ so $$r\mid \gcd (abc,d^2)=1$$, so $$r=1$$ and we are done.

The same we do for $$b$$ and $$c$$