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Let $f: \mathbb R \to \mathbb R$ be a function such as $$f(x)\cdot f(y)=f(x-y).$$ Find all possible values for $f(2018)$.

All I got is that $f(x)=0$ or $f(0)=1$ (when I put $y=0$) and $f^2(x)=f^2(y)$ (when I put $x=y$).

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  • $\begingroup$ $$x=y\implies ?, x=y=0\implies ?$$ $\endgroup$ Feb 2, 2018 at 10:04
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    $\begingroup$ Is this a question from an on-going contest? $\endgroup$
    – JRN
    Feb 2, 2018 at 10:12
  • $\begingroup$ Please use mathjax. I formatted the post for you. $\endgroup$ Feb 2, 2018 at 10:13
  • $\begingroup$ $ x = y = 0 \Longrightarrow f ( 0 ) \in \{ 0 , 1 \} $, $ y = x \Longrightarrow f ( x ) \in \{ - f ( 0 ) , f ( 0 ) \} $, $ y = \frac x 2 \& f \left( \frac x 2 \right) \ne 0 \Longrightarrow f ( x ) = 1 $. $\endgroup$ Feb 7, 2018 at 12:44

2 Answers 2

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First of all, yes, one option is that $f(x)=0$ for all $x$. That's one candidate for $f$.


Now, let's assume $f\neq 0$ (i.e., $\exists x: f(x)\neq 0$). From there you can conclude that $f(0)=1$.

Now, actually, if you plug in $x=y$, you should get

$$f(x)f(x)=f(x-x)=f(0) = 1$$

so $$f^2(x)=f(0) = 1$$

meaning that $f(x)=\cdots$ well, you can probably finish from here.

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In continuation to 5xum's answer I want highlight something that might not be that obvious. It might appear that
$$f(2018)=±1$$
But when proceeded as follows:
Put $x=2$ and $y=1$
$$f(2).f(1)=f(1)$$
$$f(2)=1$$
Now similarly putting $x=3$ and $y=2$ we obtain $f(3)=1$. Now it can easily be proven that $\forall x\in \mathbb{N}$ and $x>1$
$$f(x)=1$$ Therefore, $f(2018)$ is either $0$ or $1$ and not $-1$.

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