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.

  • 1
    $\begingroup$ $f(x)=x^2$ is a solution, if that's any help. $\endgroup$ – Ludvig Lindström May 5 at 16:30
  • $\begingroup$ Why the solution is $f(x)=x^2$ ? $\endgroup$ – 2cats May 5 at 16:34
  • $\begingroup$ Try substituting $f(a+x)$ and $f(a-x)$ for $(a+x)^2$ and $(a-x)^2$ in the equation. $\endgroup$ – Ludvig Lindström May 5 at 16:39
  • $\begingroup$ 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.) $\endgroup$ – JonathanZ supports MonicaC May 5 at 20:24

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 $$f(x)=x^2+c \ \ \ \ \forall x,c \in \mathbb{R} \ \ \ \Box$$

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