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 $(a-b)^2=f((g(a)-g(b))^2)$?



Is there any mathematical study about |a-b|=f(|g(a)-g(b)|)? or does there exist f() and g() satisfy this equation?
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.


