All solutions to $1/a+1/b=1/c$?

Question

Conjecture:
Given integers $a>b>c>1$ such that $\gcd(a,b,c)=1$. Then all the positive integer solutions to $$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$$ is given by: $c$ any number and for each divisor $d<\sqrt{c}\:$ of $\,c\,$ with $\gcd(d,c/d)=1$ $$ \left\{ \begin{array}{l} a_d=c+d^2 \\ b_d=\frac{c}{d}\cdot(\frac{c}{d}+d)=c\cdot(c+d^2)/d^2 \end{array} \right. $$ It is easy to show that $a_d,b_d,c$ all are solutions, but how to prove they are the only solutions?

Answer 1

If $a$, $b$ and $c$ are such that $a>b>c>1$ amd $\gcd(a,b,c)=1$ and $\tfrac1a+\tfrac1b=\tfrac1c$, then aso $$bc+ac=ab.\tag{1}$$ This shows that $c$ divides $ab$, and similarly $a$ divides $bc$ and $b$ divides $ac$. Let $w=\gcd(a,b)$ and $v=\gcd(a,c)$ and $u=\gcd(b,c)$. Then $u$, $v$ and $w$ are pairwise coprime positive integers with $w>v>u$, and $$a=vw,\qquad b=uw,\qquad c=uv.$$ Plugging this into $(1)$ shows that $w=u+v$, and so every solution is of the form $$a=v(u+v),\qquad b=u(u+v),\qquad c=uv,$$ with $v>u>0$ coprime. Conversely, for any two coprime positive integers $v>u$ you have $$\frac{1}{v(u+v)}+\frac{1}{u(u+v)}=\frac{1}{uv}.$$ This shows that these are precisely all solutions.

This is equivalent to your characterization; take $d=u$ and $c=uv$.