MHB Is a Harmonic Function Constant if its Square is Also Harmonic?

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Here is this week's POTW:

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Let $X$ be a domain in $\mathbb C$, and let $f : X \to \Bbb R$ be a harmonic function such that $f^2$ is harmonic. Prove $f$ is constant.

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This week's problem was correctly solved by Opalg. Here is his solution.

Spoiler

Writing $f_x,\;f_y$ for the partial derivatives, the condition for harmonicity is $f_{xx} + f_{yy} = 0.$

For the function $f^2$, $(f^2)_x = 2ff_x$ and $(f^2)_{xx} = 2(f_x)^2 + 2ff_{xx}.$ Similarly $(f^2)_{yy} = 2(f_y)^2 + 2ff_{yy}.$ Therefore $$(f^2)_{xx} + (f^2)_{yy} = 2\bigl((f_x)^2 + (f_y)^2\bigr) + 2f\bigl(f_{xx} + f_{yy}\bigr) = 2\bigl((f_x)^2 + (f_y)^2\bigr).$$ If $f^2$ is harmonic then this must be zero. But $(f_x)^2 + (f_y)^2 = 0$ implies $f_x = f_y = 0$. Therefore $f$ is constant. (Strictly speaking, if the partial derivatives vanish then $f$ is locally constant. But since a domain has to be connected it follows that $f$ must be globally constant.)