Find all triples $(f,g,h)$ of injective functions from the set of real numbers to itself satisfying \begin{align*} f(x+f(y)) &= g(x) + h(y) \\ g(x+g(y)) &= h(x) + f(y) \\ h(x+h(y)) &= f(x) + g(y) \end{align*} for all real numbers $x$ and $y$.
I have found a solution to this problem which goes like this: Let $\omega$ denote the cyclic sum with $f,g,h$ Just by switching $x$ with $y$ and summing up.
We have $$\omega (f(x+f(y)))=\omega (f(y+f(x)))\tag1$$ Now,
$$f(y+f(x))=g(y)+h(x)\tag2$$ $$g(x+g(y))=h(x)+f(y)\tag3$$
$(2)-(3)$ gives,
$$f(y+f(x))-g(x+g(y))=g(y)-f(y)\tag4$$
Now just using (4) after a rearrangement by sending $g,h$ from L.H.S. to the R.H.S. and sending $h$ from R.H.S. to L.H.S.. we get $h(x)-f(x)=h(y)-f(y)$,..... (I'm not including the rest of the solution as it's not relevant to this post) $$\underline{\textbf{My Doubt:}}$$ What does it mean by "Now just using (4) after a rearrangement by sending $g,h$ from L.H.S. to the R.H.S and sending $h$ from R.H.S. to L.H.S.." and how does it imply $$h(x)-f(x)=h(y)-f(y)$$ please can anyone show the intermediatary steps that gets you to $$h(x)-f(x)=h(y)-f(y)$$ Thanks.
I don't think that the solution is correct.
I think the "L.H.S" represents L.H.S. of $(1)$ which I think can be written as
$$f(x+f(y))+g(x+g(y))+h(x+h(y))=f(y+f(x))+g(y+g(x))+h(y+h(x))$$ so I think we have $$f(x+f(y))=f(y+f(x))+g(y+g(x))+h(y+h(x))\color{red}{-g(x+g(y))-h(x+h(y))}$$
I think that this means $$f(x+f(y))\color{red}{-h(y+h(x))}=f(y+f(x))+g(y+g(x))-g(x+g(y))-h(x+h(y))$$
I think that this means $$f(x+f(y))-h(y+h(x))=\color{red}{g(y)-f(y)}+g(y+g(x))-h(x+h(y))$$ However, I don't think that this implies $$h(x)-f(x)=h(y)-f(y)\tag5$$
The above solution is trying to show $(5)$ without using the injectivity condition.
I think that in the link, there are two other solutions in which $(5)$ appears.
However, the both solutions look incorrect.
The solution by sa2001 says "$f(x + f(y)) + f(x) + f(y) = g(x + g(y)) + h(x + h(y))$". I think this should be $f(x + f(y)) + f(x) + f(y) = \color{red}{g(y + g(x)) + h(y + h(x))}$. So, I think that the solution does not get $Q(g,x,y)$ which is equivalent to $(5)$.
The solution by sriraamster says "$B=A+C$". I think this should be $B=-A-C$ which is equivalent to $(1)$. (There is a typo. R.H.S. of $(2')$ should be $h(x + h(y)) + \color{red}{f(x + f(y))}$.) So, I think that the solution does not get $B=A+C$ which is equivalent to $(5)$.
Since the both solutions in which $(5)$ appears look incorrect, the solution in your question may have a similar error somewhere.
From the errors in the both (wrong) solutions, I think that the solution in your question may have the following two errors :
Error 1 : From $(1)$,
$$\small f(x+f(y))-h(y+h(x))=\underbrace{f(y+f(x))\color{red}-g(y+g(x))}_{}\color{red}+g(x+g(y))-h(x+h(y))\tag6$$
Error 2 : From $(6)$, $$\small f(x+f(y))-h(y+h(x))=\underbrace{g(y)-f(y)}_{}+g(x+g(y))-h(x+h(y))\tag7$$
Explanation :
$(6)$ is wrong since there are two sign errors in red.
$(7)$ is wrong since it is wrong that $f(y+f(x))-g(y+g(x))=g(y)-f(y)$. We have $(4)$ which is $f(y+f(x))-\color{red}{g(x+g(y))}=g(y)-f(y)$.
$(7)$ implies $(5)$.