Diophantine plane $ax + by + cz = 0$ with $\gcd(a,b)=\gcd(y,z)=1$.

Question

I have a constraint equation in integers that I would like to understand when there are solutions, and how to find all the solutions.

Given $a$, $b$, $c$, such that $\gcd(a,b)=1$, find $x$, $y$, $z$ with $\gcd(y,z)=1$ that satisfy: $$ax + by + cz = 0$$

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My thoughts to far:

Since $\gcd(a,b)=1$, I can use the Euclidean algorithm to solve for an $X,Y$ such that: $$aX + bY = 1, \ \ \gcd(X,Y)=1$$

There is some freedom in $X$, $Y$, such that $(X,Y)\rightarrow (X+bn,Y-an)$ for any integer $n$ will also solve that equation.

From there I'm not sure how to proceed systematically, but I can see that I can now choose $z=\pm\ 1$, and then set $x=\mp\ cX$, $y=\mp\ cY$ to solve the constraint equation. Since $|z|=1$, $\gcd(y,z)=1$. Thus all the constraints are satisfied.

So unless I made a mistake, it looks like a solution always exists. But I don't know how to determine if there are more solutions or not.

Do all solutions have $|z|=1$? Is there some simple relationship between all the solutions?

Answer 1

For any integer $z$, the solutions $(x,y)$ of $ax+by=-cz$, i.e. (with your notations) of $ax+by=-cz(aX+bY)$, are given by $$x=-czX+kb,\quad y=-czY-ka\quad(*)$$ where $k$ is any integer.

Then, $$\gcd(y,z)=\gcd(-czY-ka,z)=\gcd(ka,z)$$ so the solutions of your problem are given by:

• choose any integer $z$ coprime to $a$ ;
• choose any integer $k$ coprime to $z$ ;
• let $x,y$ be given by $(*)$.