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Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \begin{cases} f(x(1+f(x)))=f(x)^2,\\ f(x(1-f(x)))=f(x) f(-x),\\ f(x(-1+f(-x)))=f(x)f(-x). \end{cases}

I have found only that for nontrivial $f$ must be $f(0)=1$ and have no more ideas.

Leox
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    You can omit the third equation as it's obtained by replacing $-x$ for $x$ in the second equation. – Mohsen Shahriari Jul 17 '15 at 21:16
  • $f(x)=0$, $f(x)=1$ and $f(x)=x+1$ are solutions to the equation. – Mohsen Shahriari Jul 17 '15 at 21:40
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    $f(x)=0$ and $f(x)=ax+1$ are the only polynomial solutions to the equation. – Mohsen Shahriari Jul 17 '15 at 21:56
  • Might be both late and not much help, but $f : x \in \mathbb{R} \mapsto \begin{cases} 1 ,,\text{if},, x \in \mathbb{Q}\ 0 ,,\text{otherwise}\end{cases}$ can be shown to also be a solution, so getting a satisfying form of the solutions might be tricky? – Bruno B Oct 25 '22 at 18:56
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    It seems there are actually a lot of non-continuous solutions: if we let $$\begin{cases}\mathcal{A} := {A \subset \mathbb{R} \mid (0 \in A) ,,\underline{\text{and}},, (x \in A \Rightarrow -x \in A ,,\underline{\text{and}},, 2x \in A)} \ \mathcal{B} := {A \subset \mathbb{R} \mid (0 \not\in A) ,,\underline{\text{and}},, (x \in A \Rightarrow -x \not\in A ,,\underline{\text{and}},, 2x \in A)}\end{cases}$$ then the $f := _A$ for $A \in \mathcal{A} \sqcup \mathcal{B}$ are exactly the solutions $f$ such that $f(\mathbb{R}) \subset {0,1}$. – Bruno B Oct 29 '22 at 18:43

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