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I was wondered to solve the following problem:

If $f(y)=f(x)\cdot f(x-y)$ only for $x \neq y$ , what is the $f(x)$ function ?

thanks

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    It could also mean that $f(y)=f(x)f(x-y)$ is allowed to hold only if $x\ne y$ (but does not have to even then). In that case there are lots of solutions, such as $f(x)=2$ or $f(x)=e^{x+1}$ -- in fact every nowhere-zero function such that $f(0)\ne 1$.2011-09-21

3 Answers 3

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I found it.
According to assumption of problem if $y=x$ thus $f(y)\neq f(x).f(x−y)$ or $f(x)\neq f(x).f(0)$, so if $x=0$ thus $f(0)\neq f(0).f(0)=f^2(0)$ since $f(0)\neq 0,1$ $((1))$
If $y=0$ thus $f(0)=f(x)f(x-0)=f^2(x)$ so $f(0)>0$ and if $x \neq0$ thus $f(x)=\pm \sqrt{f(0)} $ $((2))$ According to assumption and $((2))$ $\Longrightarrow$ $|f(y)|=|f(x).f(x-y)|$ then $\sqrt{f(0)}=\sqrt{f(0)}.\sqrt{f(0)}=f(0)$ thus $f(0)=0 or 1$ and This is a paradox by using $((1))$, Thus $\forall x,y\in\ D_{f}$ -{0} $\Longrightarrow$ $x-y \notin D_{f}$ $\Longrightarrow$ $f(x) =$ \begin{cases} \ f(0)=c, & \ x=0, \ \pm \sqrt {c}, & \ x \ne 0, \end{cases} And $\forall x,y\in\ D_{f}$ -{0} : $x-y \notin D_{f}$

Such as $f(x) =$ \begin{cases} \ 4, & \ x=0, \ -2, & \ x \in (1,2], \end{cases}

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    everybody can not understand above answer , he can vote negative.2011-09-22
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Okay, here's one way to solve the problem.

  1. First of all, note that either $f$ is the zero function, or it is nonzero everywhere. To prove this, assume that $f(c) = 0$ for some $c$. Then for $x \neq c$, we have $f(x) = f(c) \cdot f(c-x) = 0 \cdot f(c-x) = 0$. Since $f(c) =0$ already by assumption, it follows that $f$ is the zero function.

  2. From now on, assume that $f$ is not the zero function. Then claim (1.) implies that $f(x) \neq 0$ for all $x$.

  3. For every nonzero $x$, setting $y=x/2$ gives $f(x/2)=f(x)f(x/2)$. Since $f(x/2) \neq 0$, we can cancel it to get $f(x)=1$.

  4. Finally, $f(0) = f(1) \cdot f(1 - 0) = f(1)^2 = 1$.

From (3.) and (4.), we can conclude that if $f$ is nonzero, then we must have $f(x) = 1$ for all $x$. So the only solutions are the constant functions $f(x) \equiv 0$ and $f(x) \equiv 1$.

Depending on what the OP means by "only for $x\ne y$", these two solutions are either the only solutions, or there are no solutions at all.

(We have tacitly assumed that the domain and codomain of $f$ are fields of characteristic $\ne 2$.)

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    Now it's a little bit different from my answer and so I'll keep mine.2011-09-21
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Taking $y=0$ we get $f(0)=f(x)^2$ for all $x\ne0$.

If $f(0)=0$ then $f\equiv0$.

If $f(0)\ne0$ then $f$ is never zero. In this case take $x=2y$ and get $f(y)=f(2y)f(y)$, which implies $f(2y)=1$ for all $y\ne0$. Thus $f\equiv1$.

(Thanks to Henning Makholm and Srivatsan Narayanan for helping cleaning this up.)

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    @saeed: If $f(y)=f(x)\cdot f(x−y)$ for $x \neq y$ and $f(y)\neq f(x)\cdot f(x−y)$ for $x = y$, then as pointed out several times *there are no solutions*.2011-09-21