The only solution that I could find to your equation was $f(x) = |x|$. Here is the proof (hopefully there isn't anything wrong with it): First, let us plug in $x = y = 0.$ This gives us the equality $f(f(0)) = (f(0))^2.$ Hold on to this, because we will need it later. Now let us plug in $x = y = a,$ where $a$ is some constant. This gives us that $f(f(0)) = (f(x))^2 - x^2.$ We can rearrange this to get that $f(x) = \sqrt{f(f(0)) + x^2}.$ Now that we are so close, we need to find a value for $f(f(0)).$ We'll start by plugging in $f(0)$ into the definition of the function. This gives us that $f(f(0)) = f(\sqrt{f(f(0)) + (f(0))^2}).$ Because we know that $f(f(0)) = (f(0))^2,$ we can substitute that expression for $f(f(0)) = f(\sqrt{2(f(0))^2).}$ Then we know that $f(0) = \sqrt{2}f(0).$ Clearly, the only solution to this is $f(0) = 0.$ Then our original expression for the function becomes: $\boxed{f(x) = \sqrt{x^2} = |x|}$ After typing this up, I feel as if there is something seriously wrong with this solution, but I'm not really sure what. If anyone has any ideas, please let me know. --VIPMaster

Well, you don't have injectivity so you can't deduce that $f(0)=0$ Also, let $a=f(f(0))$. Then $f(x)=\sqrt{a+x^{2}}$ is a solution by itself, but you need to prove that there doesn't exist a peace-wise function, such that $f(m)>0$ and $f(n)<0$ for some $m,n\in\mathbb{R}$. Since you only wanted to know if your solution is correct, I am only giving you the answer: partially. I haven't checked if there is any constraint to $a$, nor have I checked for peace-wise functions(it's very late where I come from), so I guess you're on your own.

sir_hoang wrote: Find all functions $f:R \to R$ satisfy the following equation $f(f(x - y)) = f(x)f(y) + f(x) - f(y) - xy$ Let $P(x,y)$ be the assertion $f(f(x-y))=f(x)f(y)+f(x)-f(y)-xy$ Let $f(0)=a$ Notice that the summand $xy$ in RHS implies that $f(x)$ can not be bounded. $P(x,0)$ $\implies$ $f(f(x))=(a+1)f(x)-a$ And so (squaring) : $f(f(x))^2=(a+1)^2f(x)^2-2a(a+1)f(x)+a^2$ $P(f(x),f(x))$ $\implies$ $f(f(x))^2=f(x)^2+f(a)$ And so $(a+1)^2f(x)^2-2a(a+1)f(x)+a^2=f(x)^2+f(a)$ And since $P(0,0)$ implies $a^2=f(a)$, we get : $af(x)((a+2)f(x)-2(a+1))=0$ Setting $x=0$ in this last equality, we get $a^2(a^2-2)=0$ and so $a=0$ or $a^2=2$ If $a^2=2$, then $af(x)((a+2)f(x)-2(a+1))=0$ implies $f(x)\in\{0,2\frac {a+1}{a+2}\}$ bounded, in contradiction with original equation. So $a=0$ and $P(x,x)$ $\implies$ $f(x)^2=x^2$ $\forall x$ Let then $x,y\notin\{0,1\}$ such that $f(x)=x$ and $f(y)=-y$ : If $f(f(x-y))=x-y$, $P(x,y)$ becomes $xy=y$, impossible If $f(f(x-y))=y-x$, $P(x,y)$ becomes $xy=x$, impossible So : either $f(x)=x$ $\forall x\ne 1$ either $f(x)=-x$ $\forall x\ne 1$ If $f(x)=x$ $\forall x\ne 1$, then $P(3,1)$ $\implies$ $2=3f(1)+3-f(1)-3$ and so $f(1)=1$ and so $f(x)=x$ $\forall x$ If $f(x)=-x$ $\forall x\ne 1$ then $P(2,0)$ $\implies$ $2=-2$, impossible Hence the unique solution : $\boxed{f(x)=x}$ $\forall x$ which indeed is a solution.

Here is my solution for this problem Solution $f(f(x - y)) = f(x) - f(y) + f(x)f(y) - xy$ Let $f(0) = a$ $P(0; 0): f(a) = a^2$ $P(a; a): a^4 - 2a^2 = 0$ $P(a; 0): f(a^2) = a^3 + a^2 - a$ $P(a^2; a^2): a^2 = (a^3 + a^2 - a) - a^4$ Then: $a^3(a^3 + 2a^2 - 2a - 2) = 0$ So: $a = 0$ or $f(0) = 0$ $P(x; 0): f(f(x)) = f(x), \forall x \in \mathbb{R}$ Then: $f(x - y) = f(x) - f(y) + f(x)f(y) - xy$ $P(x; x): f^2(x) = x^2$ So: $f(x) = x$ or $f(x) = - x$ Suppose there exist $b; c \in \mathbb{R^*}$ which satisfy $f(b) \ne b$ and $f(c) \ne - c$ Then: $f(b) = - b$ and $f(c) = c$ So: $f(- b) = f(f(b)) = f(b)$ or $f(- b) = - b$ $P(- b; c): f(- b - c) = - b - c - 2bc$ If $f(- b - c) = - b - c$ then: $2bc = 0$, conflict with $b; c \in \mathbb{R^*}$ If $f(- b - c) = b + c$: $P(b + c; c): f(b) = f(b + c) - f(c) + f(b + c)f(c) - (b + c)c$ So: $- b = f(b + c) - c + cf(b + c) - c(b + c)$ But: $f(b + c) = f(f(- b - c)) = f(- b - c) = b + c$ then: $b = 0$, conflict with $b \in \mathbb{R^*}$ So: $f(x) = x, \forall x \in \mathbb{R}$ or $f(x) = - x, \forall x \in \mathbb{R}$ Retry, we see that: $f(x) = x$, $f(x) = - x$ satisfy the problem In conclusion, we have: $f(x) = x, \forall x \in \mathbb{R}$ or $f(x) = - x, \forall x \in \mathbb{R}$