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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.

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    $\begingroup$ your proof is perfect ! $\endgroup$ – Surb Feb 16 at 18:16
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Yes this is a valid use of Bézout's identity to solve the problem.

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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 $\,0 = 1.\,$

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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 $c$

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