How to find all solutions of the optic equation $\frac{1}a+\frac{1}b = \frac{1}c$

Optic equation says for

$${\frac {1}{a}}+{\frac {1}{b}}={\frac {1}{c}}$$

All solutions in integers $a, b, c$ are given in terms of positive integer parameters $m, n, k$ by

$$a=km(m+n) , \quad b=kn(m+n), \quad c=kmn$$

where $m$ and $n$ are coprime.

How is this proved? I went through the reference, which says "A. Palmström, J. Sadier, and C. Moreau" "Ibid, 299-302". No clue what that means though.

Does anyone know how why all the solutions are covered by these formulas?

The equation $$\frac 1a+\frac 1b=\frac 1 c$$ is equivalent to $$(a+b)c=ab\quad \text {or}\quad (a-c)(b-c)=c^2$$

Note that $(ka,kb,kc)$ is a solution iff $(a,b,c)$ is, so we may assume that $\gcd(a,b,c)=1$.

If $p$ is any prime dividing $c$ then $p$ must divide either $a-c$ or $b-c$. It can't divide both, else $p$ would divide $\gcd(a,b,c)$. Thus $p^2$ divides one of $(a-c),(b-c)$. It follows that both $(a-c)$ and $(b-c)$ are perfect squares, so we may write, e.g., $(a-c)=m^2,\,(b-c)=n^2$. Then, of course, $c=mn$.

Thus every relatively prime solution is given by $$a=m^2+mn=m(m+n)\quad \quad b=n^2+mn=n(m+n)\quad \quad c=mn$$

as desired.