Solving the functional equation

Question

Find all functions $f\colon \mathbb{R}\to\mathbb{R}$ so that $f(x+y)-f(x-y)=4xy$ for all real numbers $x$ and $y$.

I put $x=y=z/2$ and got $f(z)=z^2+C$ for any real constant, but can't prove its the only one.

Thank you.

Answer 1

$\begin{array}{rcl} \forall z \in \Bbb R: f \left( \dfrac z2 + \dfrac z2 \right) - f \left( \dfrac z2 - \dfrac z2 \right) &=& 4\left( \dfrac z2\right)^2 \\ f(z) - f(0) &=& z^2 \\ f(z) &=& z^2 + C \end{array}$

where $C = f(0)$.

Now, to prove that it is consistent:

$\begin{array}{rcl} \forall x,y \in \Bbb R: f(x+y) - f(x-y) &=& [(x+y)^2 + C] - [(x-y)^2+C] \\ &=& [x^2+2xy+y^2+C]-[x^2-2xy+y^2+C] \\ &=& 4xy \end{array}$

So, we have proved that $[\forall x,y \in \Bbb R: f(x+y) - f(x-y) = 4xy] \iff [\forall z \in \Bbb R: f(z) = z^2+C]$.

Answer 2

$f(x+y)-f(x-y)=4xy $

If $f$ is differentiable, divide by $2y$ to get $\frac{f(x+y)-f(x-y)}{2y}=2x $.

Letting $y \to 0$ gives $f'(x) = 2x$ so $f(x) = x^2+c$.

This satisfies the equation for all $c$.

Remove all conditions on $f(x)$.

Let $f(x) = x^2+g(x)$. Then $4xy =(x+y)^2+g(x+y)-((x-y)^2+g(x-y)) =4xy+g(x+y)-g(x-y) $ so $g(x+y) = g(x-y) $.

Letting $x=0$ gives $g(y) = g(-y)$.

Letting $x=y$ gives $g(2x) = g(0)$.

Therefore $g(x)$ is constant.

Therefore all solutions are of the form $f(x)=x^2+c$.