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?
 A: 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.
A: If there were no limits on the right side, then it would be
$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$.
Because of the limits on the right side, the denominator has to be divisible by $a+b$, thats why one multiplies this to the denominators. This would be enough, but for saying that $m$ and $n$ are coprime, its necessary to multiply the denominators by k.
