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The problem is just as the title. It is clear that the linear function $f(x)=kx$ and $g(x)=(1/k)x$ can meet it. Is there any other function pairs f(x) and g(x) can meet this equation? or the equation $(a-b)^2=f((g(a)-g(b))^2)$?

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    $\begingroup$ You can add more restrictions and get something equivalent to Cauchy's functional equation. en.wikipedia.org/wiki/Cauchy%27s_functional_equation What does this have to do with algebraic geometry? $\endgroup$ Feb 8, 2014 at 12:53
  • $\begingroup$ I read the Cauchy's functional equation. While its form may be an example with $f(x)=g^{-1}(x)$. Sorry about the category, I'm just a newcomer, have no idea to category this question. $\endgroup$
    – firefly
    Feb 8, 2014 at 13:56

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Let me consider first a slightly different equation $|g(a)-g(b)|=f(|a-b|).$ There is no continuously differentiable function $g$ like this, except $g(x)=kx+c$.

Proof. Let $a=x$, $b=x+1$. Then we gave $g(x+1)-g(x)=\pm k,$ where $k=f(1)$, and this is for every $x$. As $g$ is continuous there must be one sign for all $x$, for example $g(x+1)-g(x)=k$. Differentiating we conclude that $g^{\prime}$ has period $1$. Repeating the same argument with $\sqrt{2}$ instead of $1$, we conclude that $g^{\prime}$ has period $\sqrt{2}$. Therefore, $g^{\prime}$ must be constant.

The same argument applies to your equation if you assume that the sets $f^{-1}(c)$ do not contain intervals. Now, for the problem to make sense, $f$ must be surjective. Therefore, for all $c$, except countably many, $f^{-1}(c)$ cannot contain an interval. Co you can find two non-commensurable values $c'$ and $c"$ whose $f$-preimages do not contain intervals. Then the previous argument shows that every $C^1$ solution $g$ must be affine.

Your second equation is the same as the first one if you replace $f$ by $f_1(x)=\sqrt{f(x^2)}$.

On the other hand, if you put no restrictions on $g$ whatsoever, one can construct crazy examples using axiom of choice.

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  • $\begingroup$ Already $g^\prime$ is periodic with period $1$ and $\sqrt{2}$, by your argument, and therefore constant. You don't need to differentiate twice. $\endgroup$
    – Yair Hayut
    Feb 8, 2014 at 18:29

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