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

Question

I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$. Let $$[f(x)]^2-4\cdot f(x^5)+3=0,\forall x\in\mathbb{R}$$

I want to prove that $f$ isn't a one-to-one function. So I suppose that $x_1,x_2\in\mathbb{R}$ with $x_1\neq x_2$ and I get:

• $x_1\neq x_2\Rightarrow x_1^5\neq x_2^5\Rightarrow f(x_1^5)\neq f(x_1^5)\Rightarrow -4\cdot f(x_1^5)+3\neq -4\cdot f(x_2^5)+3$

But with the first term I can't say that $[f(x_1)]^2\neq [f(x_2)]^2$ (because if for example $f(x_1)=1$ and $f(x_2)=-1$ I have that $(-1)^2\neq 1^2$ which isn't right.

So any ideas?

Answer 1

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.

Answer 2

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.