Integer solution of $\tfrac ax+\tfrac by=\tfrac cz$ for pairwise coprime positive integers $a$, $b$ and $c$. Given positive pairwise coprime integers $a,b,c$, how to find positive coprime integer solutions $(x,y,z)$ of
\begin{equation*}
\frac{a}{x}+\frac{b}{y}=\frac{c}{z}?
\end{equation*}
If there is no general formula or general way to solve it, then is there anything we can say about the positive integer solution set, for example it is easy to see that each $x\in\mathbb{N}$ corresponds to finitely many $y$ and $z$ values but this is not very meaningful.
For example, I calculated $(a,b,c)=(2,5,3)$: there are coprime solutions $(1,5,1)$, $(3,6,2)$, $(3,60,4)$, $(4,2,1)$, $(4,10,3)$, $(4,50,5)$ and so on.
 A: Claim: Let $a$, $b$ and $c$ be pairwise coprime integers, and $x$, $y$ and $z$ positive integers satisfying
$$\frac ax+\frac by=\frac cz.\tag{1}$$
Then there exist coprime positive integers $r$ and $s$ and positive divisors $a'$, $b'$ and $c'$ of $\gcd(a,r)$, $\gcd(b,s)$ and $\gcd(c,r+s)$, respectively, such that
$$(x,y,z)=\left(d\frac{a}{a'}\frac{s}{b'}\frac{r+s}{c'},d\frac{r}{a'}\frac{b}{b'}\frac{r+s}{c'},d\frac{r}{a'}\frac{s}{b'}\frac{c}{c'}\right).$$
for some positive integer $d$. Note that all nine fractions are in fact integers.
Conversely, every such triplet is a solution to $(1)$.

Proof. First a few reductions of the claim. If $(x,y,z)$ is a solution to $(1)$, then $(dx,dy,dz)$ is another solution for any positive integer $d$. So we obtain all solutions to $(1)$ by taking all positive integer multiples of solutions to $(1)$ with $\gcd(x,y,z)=1$.
Next note that if $(x,y,z)$ is solution to $(1)$ with $\gcd(x,y,z)=1$ and we let $e=\gcd(a,x)$, $f=\gcd(b,y)$ and $g=\gcd(c,z)$, so that
$$(a,b,c)=(ea',fb',gc')\qquad\text{ and }\qquad(x,y,z)=(ex',fy',gz'),$$
for some pairwise coprime positive integers $a'$, $b'$, $c'$, and positive integers $x'$, $y'$ and $z'$ with
$$\frac{a'}{x'}+\frac{b'}{y'}=\frac{c'}{z'},\tag{2}$$
and $\gcd(x',y',z')=1$ and moreover $\gcd(a',x')=\gcd(b',y')=\gcd(c',z')=1$.
So we obtain all solutions to $(1)$ with $\gcd(x,y,z)=1$ by appropriately scaling all solutions to $(2)$ with $\gcd(a',x)=\gcd(b',y)=\gcd(c',z)=1$, where $a'$, $b'$ and $c'$ range over the positive divisors of $a$, $b$ and $c$, respectively. Here "appropriately scaling" means taking $(x,y,z)=(\tfrac{a}{a'}x',\tfrac{b}{b'}y',\tfrac{c}{c'}z')$ for a solution $(x',y',z')$ to $(2)$.
So suppose $(x',y',z')$ is a solution to $(2)$ with $\gcd(a',x')=\gcd(b',y')=\gcd(c',z')=1$ and $\gcd(x',y',z')=1$, where $a'$, $b'$ and $c'$ are positive divisors of $a$, $b$ and $c$ respectively. Then clearing denominators shows that
$$y'z'a'+x'z'b'=x'y'c',\tag{3}$$
and hence that $x'\mid y'z'$ and $y'\mid x'z'$ and $z'\mid x'y'$. Because $\gcd(x',y',z')=1$ it follows that there exists a triplet of pairwise coprime positive integers $(u,v,w)$ such that
$$(x',y',z')=(vw,uw,uv),\tag{4}$$
and plugging this back into $(3)$ shows that
$$ua'+vb'=wc'.\tag{5}$$
In particular we see that for the trivial divisors $a'=b'=c'=1$ of $a$, $b$ and $c$ we have $u+v=w$, corresponding to the solutions of the form
$$(x,y,z)=(kav(u+v),kbu(u+v),kcuv),$$
where $k$ is any positive integer. More generally we can parametrize all solutions to $(5)$, for all positive divisors $a'$, $b'$ and $c'$ of $a$, $b$ and $c$ respectively, by first choosing $ua'$ and $vb'$ and then choosing appropriate $a'$, $b'$ and $c'$:
First note that $ua'$ and $vb'$ are coprime, because
$$\gcd(u,v)=\gcd(a',b')=1
\qquad\text{ and }\qquad
\gcd(a',x')=\gcd(b',y')=1,$$
where $x'=vw$ and $y'=uw$. So let $r$ and $s$ be coprime positive integers, and let $a'$, $b'$ and $c'$ be positive divisors of $\gcd(r,a)$, $\gcd(s,b)$ and $\gcd(r+s,c)$, respectively. Then $(u,v,w)=(\tfrac{r}{a'},\tfrac{s}{b'},\tfrac{r+s}{c'})$ is a solution to $(5)$. The corresponding solution to $(2)$ is then
$$(x',y',z')=(vw,uw,uv)=\left(\frac{s(r+s)}{b'c'},\frac{r(r+s)}{a'b'},\frac{rs}{a'b'}\right),$$
and the corresponding solutions to $(1)$ are then all positive integer multiples of
\begin{eqnarray*}
(x,y,z)&=&\left(\frac{a}{a'}x',\frac{b}{b'}y',\frac{c}{c'}z'\right)
=\left(\frac{as(r+s)}{a'b'c'},\frac{br(r+s)}{a'b'c'},\frac{crs}{a'b'c'}\right)\\
&=&\left(\frac{a}{a'}\frac{s}{b'}\frac{r+s}{c'},\frac{r}{a'}\frac{b}{b'}\frac{r+s}{c'},\frac{r}{a'}\frac{s}{b'}\frac{c}{c'}\right).
\end{eqnarray*}
Note that in this last expression, all nine fractions are in fact integers.
For the converse it is easy to verify that such a triplet indeed satisfies the equation.
