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We say that three positive integers $x,y,z$ are coprime if there exist three integers $a,b,c$ such that $$ax+by+cz=1$$

if we run $x,y,z$ over all possible coprime positive integers. Can we deduce that set of $a,b,c$ is the whole set of integers?

Safwane
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    Using your own definition, $|a|,|b|,|c|$ must (also) be coprime, since if (for example) $a < 0$, then there must be a corresponding solution with $|a| > 0$ and $x \to (-x).$ – user2661923 Jan 01 '21 at 17:39
  • @user2661923: Can you convert this into a detailed answer. – Safwane Jan 01 '21 at 17:57

1 Answers1

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Per OP's request, I am amplifying my comment into an answer. However, I am not saying that the (constraint-1) shown below is the only constraint that $|a|, |b|, |c|$ may have to satisfy. I am merely saying that (constraint-1) has to be satisfied.

We say that three positive integers $x,y,z$ are coprime if there exist three integers $a,b,c$ such that $$ax+by+cz=1$$

It is easy to establish that under the above definition, if for any given $a,b,c \in \mathbb{Z},$ there happens to exist positive integers $x,y,z$ with

$$ax+by+cz=1$$

then there must exist integers $x_1, y_1, z_1$ such that

$$(x_1 \times |a|) + (y_1 \times |b|) + (z_1 \times |c|) = 1. \tag{constraint-1}$$

Simply set
$x_1 = [x : ~\text{if}~ a > 0, ~\text{else}~ (-x)]$
$y_1 = [y : ~\text{if}~ b > 0, ~\text{else}~ (-y)]$
$z_1 = [z : ~\text{if}~ c > 0, ~\text{else}~ (-z)]$.

The only flaw in concluding that $|a|, |b|, |c|$ must therefore be coprime, is that one or two of $a,b,c$ could be equal to zero. However, if you are given that each of $a,b,c$ is non-zero, then $|a|, |b|, |c|$ must be coprime, using the OP's definition.

user2661923
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