<s>[INST] 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.
 [/INST] Because $$a^2+b^2+c^2=a^2+b^2+\frac{a^2b^2}{(a+b)^2}=\frac{(a^2+ab+b^2)^2}{(a+b)^2}.$$
I used that:
$$(a^2+ab+b^2)^2=a^4+2a^3b+3a^2b^2+2ab^3+b^4.$$
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