trouble understanding a solution to symmetric equations between three functions

Question

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 $y$.

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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.

Answer 1

I don't think that the solution is correct.

a rearrangement by sending $g,h$ from L.H.S. to the R.H.S.

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 $$f(x+f(y))=f(y+f(x))+g(y+g(x))+h(y+h(x))\color{red}{-g(x+g(y))-h(x+h(y))}$$

and sending $h$ from R.H.S. to L.H.S.

I think that this means $$f(x+f(y))\color{red}{-h(y+h(x))}=f(y+f(x))+g(y+g(x))-g(x+g(y))-h(x+h(y))$$

using (4) after a rearrangement

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 $$h(x)-f(x)=h(y)-f(y)\tag5$$

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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.

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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 $(5)$.