# How find all the functions which satisfy a given functional equation?

Determine all functions $$f:\mathbb{R}\to \mathbb{R}$$ satisfying the equation $$f(a+x)-f(a-x)=4ax$$, $$x\in \mathbb{R}$$, where any real value is available.

I came to the fact that $$f(a)=0$$, but I still do not know how to get the result to remove $$f(x)$$. Therefore, it should be confirmed that I can help here the scope of this task.

Thank you very much in advance.

• $f(x)=x^2$ is a solution, if that's any help. – Ludvig Lindström May 5 at 16:30
• Why the solution is $f(x)=x^2$ ? – 2cats May 5 at 16:34
• Try substituting $f(a+x)$ and $f(a-x)$ for $(a+x)^2$ and $(a-x)^2$ in the equation. – Ludvig Lindström May 5 at 16:39
• Are you sure the $f(a)$ has to be $0$? Could you show us what you did to prove it? (I don't think that it's true and that there's probably some mistake.) – JonathanZ supports MonicaC May 5 at 20:24

## 1 Answer

I don't know how you got to $$f(a)=0$$, but here's my humble solution:

Let $$z=a+x$$, and $$y=a-x$$, $$z+y=2a \Leftrightarrow a=\frac{z+y}{2}$$ $$x=z-a=z-\frac{z+y}{2}=\frac{z-y}{2}$$ Now, rewriting our original equation: $$f(z)-f(y)=z^2-y^2$$ Now, letting $$y=0$$ and $$f(0)=c$$ for some $$c \in \mathbb{R}$$, we get $$f(z)=z^2+c$$ Now, substituting that in the very original equation, we get $$(a+x)^2+c-((a-x)^2+c)=4ax$$ which is already an identity. Thus, we conclude that $$f(x)=x^2+c \ \ \ \ \forall x,c \in \mathbb{R} \ \ \ \Box$$