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Find all injective functions $f : \mathbb{R} \rightarrow \mathbb{R}$, such that for any $x, y \in \mathbb{R}$, they satisfy the condition: $f(xy) = f(f(x)f(y))$.

Attempt: Because f is injective, $xy=f(x)f(y)$. What now?

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    $\begingroup$ well, what if $y=1$? $\endgroup$
    – lulu
    Commented May 20 at 20:20
  • $\begingroup$ Really, you gotta try basic stuff like $x,y$ being $0,1$. $\endgroup$
    – copper.hat
    Commented May 20 at 20:29

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Let's start by checking the case when $x=0$, then we get $0=f(0)f(y)$. This means $f(0)=0$ because $f(y)=0$ would mean $f$ is not injective!

Now that we've settled the $x=0$ case, we can divide through to make,

$$\frac{f(x)}{x}\frac{f(y)}{y}=1$$

This implies that $\frac{f(x)}{x}=c$ for some constant $c$, because otherwise we could hold $y$ constant and change $x$ to break the equality.

At this point it's down hill, you simply plug back in $f(x)=cx$ to $xy=f(x)f(y)$ to find $c^2=1$. So $f(x)=x$ and $f(x)=-x$ are the only possible solutions. And we can easily check that they are.

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