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.

Let $\tilde c=-c$. Therefore we have
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{\tilde c}=0\Rightarrow ab+a\tilde c+b\tilde c=0$$
Now we can use the following identity:
$$(a+b+\tilde c)^2=a^2+b^2+\tilde c^2+2\underbrace{(ab+a\tilde c+b\tilde c)}_{=0}=a^2+b^2+\tilde c^2$$
Therefore
$$a^2+b^2+c^2=a^2+b^2+ (-\tilde c)^2=a^2+b^2+\tilde c^2=(a+b+\tilde c)^2=(a+b-c)^2\Rightarrow $$
$$a^2+b^2+c^2=(a+b-c)^2$$
With the last equation we proved what we wanted.
