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How do I show that if there are functions $f,g$ such that$$ f(g(x)+g(y))=bx+cy $$holds for all $b,c\in\mathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?

Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$ y^2+z^2=f(x,g(y-x)+g(z-x)). $$ Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?

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For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.

Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.

Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $\mathbb{R}$ into a $\mathbb{Q}$-basis of $\mathbb{R}$, which we denote as $\mathcal{B}=(g(x))_x$.

Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).

Take now $f(x,y)=0$ if $y \notin \mathcal{B} + \mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.

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  • $\begingroup$ Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know! $\endgroup$ – sam wolfe Jan 10 at 15:25
  • $\begingroup$ So $f$ is a two-variable function? $\endgroup$ – Mindlack Jan 10 at 16:35
  • $\begingroup$ In the extra case, yes. $\endgroup$ – sam wolfe Jan 10 at 16:47
  • $\begingroup$ With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction. $\endgroup$ – Mindlack Jan 10 at 16:47
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    $\begingroup$ What I did prove is that there is no contradiction (at least if the axiom of choice is granted). $\endgroup$ – Mindlack Jan 17 at 7:35

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