# Does the system $xy = ab, G(x)+G(y)=G(a)+G(b)$ always have exactly two solutions if $G$ is continuous and injective?

If $$f$$ and $$g$$ are commutative operations $$\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R},$$ then for any constants $$a,b \in \mathbb{R}$$, the system of equations $$f(x,y) = f(a,b), \qquad g(x,y) = g(a,b)$$ must include $$(x,y) = (a,b)$$ and $$(x,y) = (b,a)$$ among its solutions.

I conjecture that if $$f$$ is given by $$f(x,y) = xy$$, and if $$g$$ can be expressed as $$g(x,y) = G(x) + G(y)$$ for some $$G : \mathbb{R} \rightarrow \mathbb{R}$$ that's continuous and injective, then the above system has only these solutions.

The injectivity requirement in particular is designed in particular to block $$G(x) = x^2,$$ which would break the theorem if it were allowed, while allowing the choices $$G(x) = x^3$$ and $$G(x) = \mathrm{tan}^{-1}(x),$$ which seem to be consistent with the theorem (based on graphical considerations).

Here's a graphical example illustrating the case $$a = 2, b = 3, G(x) = x^3$$:

Question. Is this true? If not, what would be a counterexample? If so, how might we prove it?

• The term "commutative function" is rather unusual : Do you mean $f(x,y)=f(y,x)$ for all $x,y$ ? – Jean Marie May 9 at 11:57
• @JeanMarie, yep. I've just changed it. – goblin May 9 at 11:59

It's not true.

Define

$$G(x)= \begin{cases} x, & \text{for } x \le 4 \\ \frac12x+2, & \text{for } x > 4. \\ \end{cases}$$

This is a monotonically increasing, continous (check $$x=4$$ on both 'branches') function, thus injective.

Especially note that

$$G(2)=2, G(3)=3, G(4)=4, G(6)=5$$

and thus $$g(2,6)=7=g(3,4)$$, and of course $$2\times6=3\times4$$.