Find all functions such that $f: R \longrightarrow R, \forall x \in R, f(x)f(x^2-1)=\sin(x) $.

Question

Find all functions such that

$$f: \mathbb{R} \longrightarrow \mathbb{R} \\ f(x)f(x^2-1)=\sin(x), \quad\forall x \in \mathbb{R}$$

That is a difficult problem for me. Help me please.

Answer 1

Take first $x=\pi$. As $\sin(\pi)=0$, you have $f(\pi)=0$ or $f(\pi^2-1)=0$

1) If $f(\pi )=0$, we take $x=\sqrt{1+\pi}$. Then $x^2-1=\pi$, and hence $\sin(x)=0$. This imply that there exists $n\in \mathbb{N}$, $n\not = 0$, such that $\sqrt{1+\pi}=n\pi$. Hence $n^2\pi^2=1+\pi$, a contradiction as $\pi$ is transcendental.

2) If $f(\pi^2-1)=0$, then take $x=\pi^2-1$. We get $\sin(\pi^2-1)=0$, and hence there exists $m\in \mathbb{Z}$ such that $\pi^2-1=m\pi$, again a contradiction.

Hence such a function does not exists.

Answer 2

Here is a shorter solution.

Take $x=\frac{1-\sqrt{5}}{2}$. The L.H.S is clearly $f(x)^2$, since $x^2-1=x$.

However, note that $\sin x$ is negative, since $-\pi < \frac{1-\sqrt{5}}{2} < 0$.

Therefore, we have L.H.S nonnegative and R.H.S negative, contradiction.

The conclusion - there are no such functions.