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It is rather obvious that for any 3 coprime integers $x,y,z$ there exist 3 non-zero integers $a,b,c$ such that: $$ax+by+cz=0$$ Any simple argument to prove it?

Thomas Andrews
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    How about $a=b=c=0$? – Servaes Aug 27 '15 at 00:37
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    Take $a=yz$, $b=xz$,$c=-2xy$ (if you want them all to be non-zero). – lulu Aug 27 '15 at 00:40
  • @Servaes, thanks. I edited the question. Obviously, i was looking for non trivial solutions. –  Aug 27 '15 at 00:40
  • @lulu, did you mean take $c=-2xy$? –  Aug 27 '15 at 00:42
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    Indeed I did, and it is now corrected. – lulu Aug 27 '15 at 00:43
  • I totally agree. Can this approach be considered as a valid proof? I am just asking> –  Aug 27 '15 at 00:45
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    Of course it is valid! Constructive proofs of existence (that is, proofs devised by explicitly building the desired solution) are great. Too rare, honestly, but such is math. – lulu Aug 27 '15 at 00:47
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    As I note below, it isn't obvious if $x=y=0$ and $z=1$, then the only solutions have $c=0$. – Thomas Andrews Aug 27 '15 at 00:49
  • @ThomasAndrews Good point, and correct, of course. I made the (unsupported) assumption that $xyz≠0$. Your solution is more general. – lulu Aug 27 '15 at 00:50
  • I would like to stress that the solutions to this equation are not trivial, since they represent the solutions of ax + by = cz. – theREALyumdub Aug 27 '15 at 00:58
  • By the way, picking $a,b,c$ to be the variables you are solving for is a bit strange. Sure, it is equivalent, but it makes it easier to think in terms of patterns mathematicians know if you had phrased the question: "If $a,b,c$ are coprime integers, then there exists non-zero $x,y,z$ so that $ax+by+cz=0$." $x,y,z$ are almost always the variables you are trying to solve for in Diophantine equations. – Thomas Andrews Aug 27 '15 at 01:13

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Assume $z\neq 0$. One of them needs to be for $x,y,z$ to be coprime.

Find $u,v\neq 0$ so that $ux+vy\neq 0$. Then let $a=zu,b=zv,c=-(ux+vy)$.

There is always such $u,v$ unless one of $x,y$ is zero. Assume $x=0$ and $y\neq 0$ then choose $a=1,b=z,c=-y$.

If $z=1$ and $x=y=0$, then you can't find a solution with non-zero $c$.

You don't really need coprime, of course. Just that at least $2$ of $x,y,z$ are non-zero. Also, if all three are zero, you can trivially solve with $a=b=c=1$.

Thomas Andrews
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  • Just picking your brains for a minute, in general, it is always possible to write any integer $z$ in function of 2 coprime integers $x,y$ as followed:$$z=ax+by$$. Following your argument, $u,v$ are the Bezout coefficients? –  Aug 27 '15 at 00:54
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    We don't know that $x,y$ are coprime, only that $x,y,z$ are coprime. If the problem meant pair-wise coprime, then we could eliminate the case where $x=y=0$. @Ramunjndscpl https://en.wikipedia.org/wiki/Coprime_integers#Coprimality_in_sets – Thomas Andrews Aug 27 '15 at 00:55
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    No, if $x=1$ and $y=6$ then $x\cdot 1 + y\cdot 0=1$ is a solution of the Bezout identity, but we want a non-zero solution $(u,v)$. But, as I say, you don't know that $x,y$ are coprime. and It doesn't matter. If $y\neq 0$, either $x+y\neq 0$ or $x-y\neq 0$, so you know there is a solution with $(u,v)\in{(1,1),(-1,1),(1,-1)}$. – Thomas Andrews Aug 27 '15 at 00:59
  • Indeed, it is absolutely trivial to prove that one of $x+y$ or $x+2y$ is non-zero, unless both $x=0$ and $y=0$. So you can be sure you can use $(u,v)=(1,1)$ or $(u,v)=(1,2)$. – Thomas Andrews Aug 27 '15 at 01:11
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I am going to change the language of the problem for clarity: Let a, b and c be the constants that satisfy the property of being coprime. You asked what the solutions are to satisfy:

$$ ax + by = cz $$

For variables x, y, and z in $ \mathbb Z $

Note that in your form, one merely needs make z the opposite sign to yield a solution, then flip the variable names as I did.

We say that $ ax + by $ is a linear combination of cz. If the gcd(x,y) | cz, we have an easy case to do all the solutions with the extended euclidean algorithm. Note that you specified x and y coprime, so the gcd(x, y) = 1. See extended euclidean algorithm.

Otherwise, there are infinitely many solutions for large enough cz. This is evident in the fact that if x and y are coprime, then there exists a linear combination for any value larger than xy - x - y, a fact that can be found here. In fact this is almost a duplicate problem.

If x and y are not coprime, divide both sides by their greatest common divisor and adjust z to be appropriate for the problem. Tada! Now they are coprime.

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theREALyumdub
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$$ax+by+cz=0$$

Let $(a,b,c)=m$ $$a=mA$$ $$b=mB$$ $$c=mC$$ where $(A,B,C)=1$ then $$Ax+By+Cz=0$$ Then for all integers $u,v,w$, we have: $$x=Bw-Cv$$ $$y=Cu-Aw$$ $$z=Av-Bu$$ We confirm that $$A(Bw-Cv)+B(Cu-Aw)+C(Av-Bu)= ABw-ACv+BCu-ABw+ACv-BCu= 0$$