parmenides51 wrote: Find all functions $f$ that satisfy $$[f(x)] \cdot \left[f\left(\sqrt[3]{\frac{2+x}{2-x}}\right) \right]^2=\frac{x^3}{4}$$ I suppose that notation $[.]$ means floor function $\lfloor.\rfloor$ I suppose that missing domain of function is $\mathbb R$ I suppose that missing codomain of function is $\mathbb R$ I suppose that missing domain of functional equation is $\forall x\in\mathbb R$ If so, $\boxed{\text{No such function}}$ Just set $x=1$ and $LHS\in\mathbb Z$ while $RHS\notin\mathbb Z$

parmenides51 wrote: Find all functions $f$ that satisfy $$[f(x)] \cdot \left[f\left(\sqrt[3]{\frac{2+x}{2-x}}\right) \right]^2=\frac{x^3}{4}$$ Notation [...] in the official document was used only to avoid parenthesis inside parenthesis . I changed it afterwards to avoid misunderstandings.

parmenides51 wrote: Find all functions $f$ that satisfy $$f(x) \cdot \left(f\left(\sqrt[3]{\frac{2+x}{2-x}}\right) \right)^2=\frac{x^3}{4}$$ I suppose that missing domain of function is $\mathbb R$ I suppose that missing codomain of function is $\mathbb R$ I suppose that missing domain of functional equation is $\forall x\in\mathbb R$ If so, $\boxed{\text{No such function}}$ Just set $x=2$ and $LHS$ is undefined while $RHS=2$

looking at the solution they suppose (in the background) that $f:R -\{2\} \to R$ (e.g. $x \ne 2$) and they don't care whether everything under the third radical is positive or not, I do not know how they behave to such radicals there

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parmenides51 wrote: Find all functions $f:R -\{2\} \to R$ that satisfy for all $x \ne 2$ $$f(x) \cdot \left(f\left(\sqrt[3]{\frac{2+x}{2-x}}\right) \right)^2=\frac{x^3}{4}$$ Setting $x=0$, we get $f(0)f(1)^2=0$ If $f(0)=0$, setting $x=-2$, we get $0=-2$ If $f(1)=0$, setting $x=1$, we get $0=\frac 14$ And so $\boxed{\text{No such function}}$

their result is

so probably they suppose $x\ne 0$ (just added it above)

parmenides51 wrote: Find all functions $f:R -\{0,2\} \to R$ that satisfy for all $x \ne 0,2$ $$f(x) \cdot \left(f\left(\sqrt[3]{\frac{2+x}{2-x}}\right) \right)^2=\frac{x^3}{4}$$ Just set $x=-2$ and LHS becomes undefined, and so equation wrong. And so $\boxed{\text{No such function}}$

I send the official solution (it was not chosen on the contest)

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