1

$$ f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy $$

$$ P(x, 0): f(0)=0 \\ P(0, y): f(f(y)-f(-y))=-y^2\\ P(x, x): f(f(2x))=x^2 \\ P(x, -x): f(-f(2x))=-x^2 \\ P(-x, x): f(-f(-2x))=-x^2 \\ P(-x, -x): f(f(-2x))=x^2 \\ $$ I don't have an idea of the function $f$, but I think we can show that $f$ is an even function. $$ P(f(x), f(y)): f(f(f(x)+f(y))-\frac{x^2-y^2}{4})=f(x)f(y) \\ P(f(y), f(x)): f(f(f(x)+f(y))+\frac{x^2-y^2}{4})=f(x)f(y) \\ P(f(x), -f(y)): f(\frac {x^2-y^2}{4} - f(f(x)+f(y)))=-f(x)f(y) \\ P(f(y), -f(x)): f(-\frac {x^2-y^2}{4} - f(f(x)+f(y)))=-f(x)f(y) \\ $$ Please give a perfect process of getting the functions.

Mohsen Shahriari
  • 6,718
RDK
  • 2,623
  • 1
  • 8
  • 30

1 Answers1

3

Let us rewrite the problem as $f(f(a)-f(b)) = \frac{a^2-b^2}{4}$.

Suppose that there exists a solution;

Claim $1)$ If $f(u) = f(v)$ then $u = \pm v$.

Proof; As you noted (taking $a=b$) we have that $f(0) = 0$. Now

$$0 = f(0) = f(f(u)-f(v)) = \frac{u^2-v^2}{4} \Rightarrow u = \pm v.$$

Claim $2)$ $f(u) = f(-u)$.

Proof; Set $s = f(u)-f(-u)$. We have

$$f(s) = f(f(u)-f(-u)) = 0$$

From $f(0) = 0$ and Claim $1)$ we must have that $s = \pm 0 = 0$.

Now for all $a \in \mathbb{R}$ we have $$\frac{a^2}{4} = \frac{a^2-0^2}{4} = f(f(a)-f(0)) = f(f(a))$$

$$-\frac{a^2}{4} = \frac{0^2-a^2}{4} = f(f(0)-f(a)) = f(-f(a)) $$

which contradicts claim 2) when we consider $a \neq 0$.

acreativename
  • 3,376