$f:\mathbb{R}\rightarrow\mathbb{R}$ is not one-to-one if $[f(x)]^2-4\cdot f(x^5)+3=0,\forall x\in\mathbb{R}$?

We have that $f(1),f(-1)$ and $f(0)$ are solutions of the quadratic equation $$t^2-4t+3=0.$$ Thus

$$f(1),f(-1),f(0)\in\{1,3\}$$ which shows that $f$ is not one-to-one.


Consider the cases $x=0$, $x=1$ and $x=-1$. Since for these three options $x^5=x$, you can rewrite the condition as $(f(x)-1)(f(x)-3)=0$. Hence at least two of the three values above are mapped to the same real.

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