Find all $f : \mathbb R\to\mathbb R$ such that $f(f(x+y)) = f(x+y) + f(x)f(y) -xy$

Question

Find all $f : \mathbb R\to\mathbb R$ such that

$$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$ for all reals $x,y.$ (Belarusian Mathematical Olympiad-1995)

My answer:

Consider $f(0) = c. ...(i)$

Let $x,y = 0$ at first.

$$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$ $$\implies f(f(0+0)) = f(0+0) + f(0)f(0) -0 \times 0$$ $$\implies f(f(0)) = f(0) + f(0)f(0) + 0$$ $$\implies f(c) = c + c \times c + 0$$ $$\implies f(c) = c + c^2$$ $$\implies f(c) - c = c^2 ...(ii)$$

Now, let $y = -x.$

Therefore, $$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$ $$\implies f(f(x-x)) = f(x-x) + f(x)f(y) -(x\times -x)$$ $$\implies f(f(0)) = f(0) + f(x)f(-x) + x^2$$ $$\implies f(c) = c + f(x)f(-x) + x^2$$ $$\implies f(c)-c = f(x)f(-x) + x^2$$

From $..(ii),$

Therefore, $$c^2 = f(x)f(-x) + x^2$$

Now, let $x = c$

$$Therefore, c^2 = f(x)f(-x) + c^2$$ $$\implies c^2 - c^2 = f(c)f(-c)$$ $$\implies 0 = f(c)f(-c)$$

Hence,

$f(c) = 0 ...(iii)$

$f(-c) = 0 ...(iv)$

Multiplying $(i) and (iii),$ $$f(0) \times f(c) = c \times 0$$ $$\implies f(0) \times f(c) = 0$$ $$\implies f(0) = 0$$ $$\implies c = 0$$

Similarly, Multiplying $(i) and (iv)$

$\implies c = 0$

Hence $c = 0.$

But $f(c) = 0,$

Therefore, $f(c) = c$

I can't go further than that can anyone help me out.

Answer 1

$$f(f(x+y))=f(x+y)+f(x)f(y)-xy$$

As mentioned above, I have assumed $f(0)=c$ for some real $c$.

• $(x,y)\equiv(c,-c)$ $$f(c)=c+f(c)f(-c)+c^2$$
• $(x,y)\equiv(0,0)$ $$f(c)=c+c^2$$ The above equations imply that $f(c)f(-c)=0$. Hence, there exists some real $k$ such that $f(k)=0$.
• $(x,y)\equiv(k,0)$ $$f(f(k+0))=f(k+0)+f(k)f(0)-k\cdot0\implies c=0\implies f(0)=0 $$
• $(x,y)\equiv(x,0)$ $$f(f(x))=f(x)\implies f(f(x+y))=f(x+y)\implies f(x)=x\cdot \frac{y}{f(y)}$$ If $f(x)=0$ for all real $x$, we have, $xy=0$ for all reals,which is a contradiction. Hence there exists some real $p$ such that $f(p)=q$ and $q\ne 0$.
• $(x,y)\equiv(x,p)$ $$f(f(x+p))=f(x+p)\implies f\left((x+p)\cdot\frac{p}{q}\right)=(x+p)\cdot\frac{p}{q} \implies p=q$$ Therefore, $\boxed{f(x)=x}$ for all real $x$. Substitute this back into the original F.E to verify the result.