<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] Below we show how to simply derive the value of  $\,\color{#90f}x,$ using only $\rm\color{#0a0}{\textit{difference of squares}}\,$ factorization
$$\begin{align} 
 a^2+b^2+\color{#c00}c^2\, &=\, \color{#90f}{x^2},\ \ \ {\rm let}\ \ \ X = (a+b) x\\
\smash[t]{\overset{\!\!\!\large \times\ (a+b)^2}\iff}\ (a+b)^2(a^2+b^2) + (\color{#c00}{ab})^2 &=\, X^2\ \ \ \ {\rm by}\ \ \ \left[\,(a+b)\color{#c00}c = \color{#c00}{ab}\,\right]^2\\[.3em]
\iff\qquad\quad\ \ (a+b)^2(a^2+b^2)  &=\, \color{#0a0}{X^2 - (\color{c00}{ab})^2}\ \\[.3em]
\smash[b]{\iff\ \, (\underbrace{a^2+b^2+2ab}_{})(\underbrace{a^2+b^2}_{})} &= {(X-ab)(X+ab)}\\
  X^{\phantom{|^{|}}}\!\!\!+ab\ \ \ \ \ \ \ \ \  X-ab,\ \!\! &\ \ \  X = a^2+b^2+ab
\end{align}\qquad\qquad\quad\ $$
Thus $\ \color{#90f}x = \dfrac{X}{a+b} = \dfrac{a^2+b^2+ab}{a+b}\ = \dfrac{(a+b)^2-ab}{a+b} = a+b-c\ $ is a solution.
Remark $ $ With that insight we can easily derive the key identity at the heart of it, which is simply the symmetric polynomial Newton identity $\ p_2 = \color{#0a0}{p_1^2} - 2\, \color{#c00}{e_2},\ $ with $\,C\to -C,\,$ i.e.
$$A^2+B^2+C^2 = \color{#0a0}{(A+B-C)^2} + 2(\color{#c00}{(A+B)C-AB})\qquad$$
hence we infer that $\ A^2+B^2+C^2\ $ is a $\,\rm\color{#0a0}{square},\,$ when $\,\color{#c00}{(A+B)C = AB}$.
Proof $ $ expanding $\,\color{#0a0}{((A+B)-C)^2}$ yields RHS $= (A+B)^2\!-2AB+C^2 = $ LHS
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