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During my mathematical musings I encountered the following functional equation : denote by $L$ the set of all functions ${\mathbb Z}^2 \to {\mathbb C}$ satisfying

$$ \begin{array}{cl} &f(x+a,y+b)+f(x+b,y+c)+f(x+c,y+a) \\ =& f(x+a,y+c)+f(x+b,y+a)+f(x+c,y+b) \end{array}\tag{1} $$

for any $x,y,a,b,c\in{\mathbb Z}$. The following three classes of functions are clearly solutions :

$$ f(x,y)=g(x), \ f(x,y)=h(y), \ f(x,y)=i(x+y) \tag{2} $$

where $g,h,i$ are arbitrary functions ${\mathbb Z}\to {\mathbb C}$. Are there other solutions besides linear combinations of those ?

Ewan Delanoy
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  • I think that $f(x,y)=xy$ is possible. – HK Lee Oct 04 '14 at 16:15
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    @HeeKwonLee It is possible but it is not a new solution : $xy=\frac{(x+y)^2-x^2-y^2}{2}$ – Ewan Delanoy Oct 04 '14 at 16:23
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    Here $f$ is bilinear ? $f(x+z,y)=f(x,y)+f(z,y)$ and so on. What is meaning of linear ? – HK Lee Oct 04 '14 at 16:26
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    @HeeKwonLee No, an individual solution $f$ is not linear or bilinear in any way ; it is the equation that is linear in $f$, so that $\lambda f_1+\mu f_2$ is a solution if $f_1$ and $f_2$ are – Ewan Delanoy Oct 04 '14 at 16:27
  • [hint to reduce the problem] Clearly any function such that $f(x,y) = f(y,x)$ solves the equation; for example: $i(x+y)$ and $j(x,y)$. If one could show that any function that does not change when commuting the arguments can be written as a linear combination of $g(x)$, $h(y)$ and $i(x+y)$, the problem would be reduced to check solutions that depends both on $x$ and $y$, and that $f(x,y)\neq f(y,x).$ So, $f(x,y)=f(y,x) \Rightarrow f(x,y) = \alpha,g(x) + \beta,h(y),+\gamma,i(x+y);;?$ – Daniel Cunha Jul 07 '17 at 12:17
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    @DanielCunha I doubt your claim "Clearly any function such that f(x,y)=f(y,x) solves the equation". I tried f(x,y)=(x-y)^4 and the left side minus the right side is = 24(a-b)b-c)(c-a)(x-y) which is not zero. – Somos Apr 28 '18 at 22:02
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    Any quadratic form satisfies the constraint by the observed facts. It is easy to see that a cubic form works if and only if it is a linear combination of $x^3$, $y^3$ and $(x+y)^3$. How far did you check? – Jyrki Lahtonen Apr 15 '21 at 08:33
  • The same for quartics. – Jyrki Lahtonen Apr 15 '21 at 08:40
  • @JyrkiLahtonen Good point, I hadn't thought of it. A natural sub-question is indeed to describe the polynomial solutions. – Ewan Delanoy Apr 15 '21 at 09:39

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