Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$ 
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$  Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers. 

It can probably be solved by a quick factoring trick, but I really can’t figure it out.

So we have $c(a+b)=ab$. Now \begin{eqnarray}a^2+b^2+c^2 &=&\underbrace{a^2+\color{red}{2ab}+b^2}\color{red}{-2c(a+b)}+c^2\\ 
&=&(a+b)^2-2c(a+b)+c^2\\ 
&=& (a+b-c)^2
\end{eqnarray}
