Put $y=0$ to get $f(x)f(0)=f(x)$ see $f\equiv 0$ never works hence $f(0)=1$. Now $x+y=0$ gives $f(x)f(-x)=1-x^2$ implying either $f(1)=0$ or $f(-1)=0$. Now for the first case $y=1$ gives $f(x+1)=-x\implies f(x)=1-x$ and for the second case we have putting $y=-1$ gives $f(x-1)=x\implies f(x)=1+x$ so these are all the solutions.

From $P(x,-x)$ we get $f(x)f(-x)=-x^2+1$ From $P(x,x)$ we get $f(x)^2=f(2x)+x^2$ From $P(2x,-x)$ we get $f(2x)f(-x)=f(x)-2x^2$ By solving this set of equations we get $(f(x)+1)^2=x^2$ which implies $f(x)=x+1$ or $f(x)=-x+1$ for every $x$. We can easily check that only one of them can occur.

Let $ P(x,y) $ be the assertion $ f(x)f(y)=f(x+y)+xy $ $ P(x,0) \Rightarrow f(0)=1 $ (This is because that $ f $ is not a constant function) $ P(1,-1) \Rightarrow f(1)=0 $ or $ f(-1)=0 $ Case $ 1. f(1)=0 $ $ P(x,1) \Rightarrow f(x)=-x+1 $ Case $ 2. f(-1)=0 $ $ P(x,-1) \Rightarrow f(x)=x+1 $ So,$ \boxed{f(x)=x+1,-x+1\text{ }\forall x} $

$P(x,0) : f(x)f(0)=f(x), \forall x \in \mathbb{R}$. If $f(0)=0$, then $f(x)=0, \forall x \in \mathbb{R}$ -- which is not a solution! $\therefore f(0)=1$ ${P(1,-1) : f(1)f(-1) = 0 \implies 0 \in \{f(1), f(-1)}$ $f(1)=0 \implies P(x-1,1) : f(x) =1-x, \forall x \in \mathbb{R}$. $f(-1)=0 \implies P(x+1,-1) : f(x) =1+x, \forall x \in \mathbb{R}$.

Olemissmath wrote: Find all functions f: $R \to R$ such that : $f(x)*f(y)=f(x+y)+xy$ for all $x,y \in R$. If $y = 0$ $\implies$ $f(x)f(0) = f(x)$.If $f(x) = 0$ $\implies$ $xy = 0$ $\forall$ $x,y \in R$,contradiction $\implies$ $f(0) = 1$. If $x = 1$ and $y = -1$ $\implies$ $f(1)f(-1)=0$. If $y = -1$ $\implies$ $f(x)f(-1) = f(x-1)-x$. If $y = 1$ $\implies$ $f(x)f(1) = f(x+1)+x$. If $f(1) = 0$ $\implies$ $f(x+1) = -x$ $\implies$ $f(x) = 1 - x$. If $f(-1) = 0$ $\implies$ $f(x-1) = x$ $\implies$ $f(x) = 1 + x$. So the solution are $f(x) = 1 + x$ or $f(x) = 1 - x$.

Let $P(x,y)$ be the assertion. Then $P(0,0)\implies f(0)=1$ or $f(0)=0$. If $f(0)=0, P(x,0)\implies f\equiv 0$. If $f(0)=1, P(x,-x)\implies 1-x^2=f(x)f(-x)\implies$ either $f(1)=0$ or $f(-1)=0$. Now, if $f(1)=0, P(x,1)\implies f(x)=1-x$ and if $f(-1)=0, P(x,-1)\implies f(x)=x+1$.

From $f(x)f(c)=f(x+c)+cx $, why result $f(x)=x+1$?

Feridimo wrote:

TuZo wrote: From $f(x)f(c)=f(x+c)+cx $, why result $f(x)=x+1$? $ y=c,f(c)=const \implies \frac{(1-x^2)(1-c^2)}{f(x)f(c)}=\frac{1-(x+c)^2}{f(x+c)}+cx $ $ and $ $f(x)f(c)=f(x+c)+cx $ $ f(x)=a,f(x+c)=b $

You find $ a $ in this equation and replace it in the equation.

Answer: $f(x) = 1 + x \quad \text{or} \quad 1 - x$ Proof: Let $P(x,y)$ denote the given assertion. $P(x,0) \to f(x) \cdot f(0) = f(x) \implies f(0) = 1$ ($f(x) = 0$ doesn't work) $P(1,-1) \to f(1) \cdot f(-1) = f(0) - 1 = 0$ $\implies f(1) = 0 \quad \text{or} \quad f(-1) = 0$ Case 1: $f(1) = 0$ $P(x-1,1) \to f(x-1)\cdot f(1) = f(x) + x-1 = 0 \implies \boxed{f(x) = 1-x}$ Case 1: $f(-1) = 0$ $P(x+1,-1) \to f(x+1)\cdot f(-1) = f(x) - x - 1 = 0 \implies \boxed{f(x) = 1 + x}$

$P(x,0)\Rightarrow f(x)f(0)=f(x)$ so either $f$ is constant (no solutions) or $f(0)=1$. $P(1,-1)\Rightarrow f(1)f(-1)=0$ Case 1: $f(1)=0$ $P(x-1,1)\Rightarrow\boxed{f(x)=-x+1}$ Case 2: $f(-1)=0$ $P(x+1,-1)\Rightarrow\boxed{f(x)=x+1}$ Both of which do work.

another sol P(x,0) => f(x)=constan or f(0)=1 now put 1 and -1 => f(1)f(-1) = 0 so f(1)=0 or f(-1)=0 if f(a)=0 put x-a,a => f(x)+(x-a)a=0 now put x=0 from f(0)=1 => 1-a^2=0 => a=1 or -1 => assume f(1)=0 now again put x-1,1 => f(x)=1-x second case ( f(-1)=0) is same

Clearly $f$ is not constant. Let $P(x,y)$ the assertion of the given F.E. $P(x,0)$ $$f(0)=1$$$P(1,-1)$ $$f(1)f(-1)=0 \implies f(1)=0 \; \text{or} \; f(-1)=0$$If $f(1)=0$ $P(x-1,1)$ $$f(x)=1-x$$If $f(-1)=0$ $P(x+1,-1)$ $f(x)=x+1$ Thus the solutions are $f(x)=x+1$ and $f(x)=1-x$ thus we are done

$P(x,0): f(x)f(0)=f(x)$. If $f(0)\ne1$, then $f\equiv0$, which isn't a solution. So $f(0)=1$. $P(1,-1): f(1)f(-1)=0$. Case 1: $f(1)=0$. $P(x,1): f(x+1)=-x\implies \boxed{f(x)=-x+1 \forall x\in\mathbb{R}}$, which works. Case 2: $f(-1)=0$. $P(x,-1): f(x-1)=x \implies \boxed{f(x)=x+1 \forall x\in\mathbb{R}}$, which works. Since we have exhausted all cases, those are the only solutions.

This is BrMO Round 1 2009/2010 Problem 5.

$$f(x)f(y)=f(x+y)+xy \ldots (\alpha)$$In $(\alpha) x=y=0:$ $$f(0)^2=f(0)$$$$ \Rightarrow f(0)\in \{ 0,1\}$$If $f(0)=0:$ In $(\alpha) y=0:$ $$\Rightarrow f(x)=0, \forall x \in \mathbb{R}$$Replacing in $(\alpha)$: $$\Rightarrow xy=0, \forall x,y\in \mathbb{R} (\Rightarrow \Leftarrow)$$$$\Rightarrow f(0)=1$$In $(\alpha) x=1,y=-1:$ $$\Rightarrow f(1)f(-1)=0$$If $f(1)=0:$ In $(\alpha) x=1:$ $$\Rightarrow f(x)=1-x$$If $f(-1)=0:$ In $(\alpha) x=-1:$ $$\Rightarrow f(x)=x+1$$$$\Rightarrow f(x)=\pm x+1 _\blacksquare$$

Subbing $y=0$ gives $(f(0)-1)f(x)=0$, so $f(0) = 1$ since clearly $f(x)=0$ is not a valid function. Letting $x=-y$ gives $f(x)f(-x) = 1 - x^2$, and the case $x=1$ gives us $f(1)f(-1) = 0$. Thus, either $f(1)$ or $f(-1)$ is $0$. Plugging in $y = 1$ and $y=-1$ gives $f(x)f(1) = f(x+1) + x$ and $f(x)f(-1) = f(x-1) - x$ respectively. If $f(1) = 0$ we get the function $\boxed{f(x) = 1-x}$, and if $f(-1) = 0$ we get the function $\boxed{f(x) = 1+x}$. Substituting these functions back into our original equation shows that they both work.