<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] Let $\dfrac1a=A$ etc.
$\implies A+B=C$
$$a^2+b^2+c^2=\dfrac{(AB)^2+(BC)^2+(CA)^2}{(ABC)^2}$$
$$(AB)^2+(BC)^2+(CA)^2$$
$$=(AB)^2+C^2(A^2+B^2)$$
$$=(AB)^2+(A+B)^2(A^2+B^2)\text{ if }C=\pm(A+B)$$
$$=(AB)^2+(A^2+B^2+2AB)(A^2+B^2)$$
$$=(AB)^2+(A^2+B^2)^2+2\cdot AB\cdot(A^2+B^2)=?$$
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