Cauchy-like functional equation

Question

I came across this question from some Olympiad training material, which has a strong likeness to Cauchy functional equations:

Find all functions $f : \mathbb{Q} \times \mathbb{Q} \mapsto \mathbb{Q}$ such that $f(x,y) + f(y,z) + f(z,x) = f(0,x+y+z)$

My questions:

• How to get rid of the extra dimension, it looks like $f(x,y)=f(0,y)=g(y)$ but how it can be shown

• How to reduce the given condition from 3 variables to 2.

Answer 1

Working in $\mathbb Q$ is a strong hint that you must divide and conquer.

As proven by Bruno B in the comments, $f(x,0)=0$ for all $x$. Then:

1. Plug $(x,y,0)$ to get $f(x,y)=f(0,x+y)-f(0,x)$. From this point on, forget about $f$ altogether. Let's just work with $g(x)=f(0,x)$.
2. By expressing $f$ via $g$, out equation turns into $g(x+y)+g(x+z)+g(y+z)-g(x)-g(y)-g(z)=g(x+y+z)$. This kind of says that the third order finite difference derivative of $g$ is $0$, so $g(x)$ must be a quadratic polynomial. But if we feel the urge to be really rigorous, proceed to the following steps.
3. Plug $(x,x,x)$ to get $g(3x)=3g(2x)-3g(x)$
4. Plug $(2x,x,x)$ to get $g(4x)=2g(3x)+g(2x)-g(2x)-2g(x)=6g(2x)-8g(x)$
5. Plug $(3x,x,x)$ to get $g(5x)=2g(4x)+g(2x)-g(3x)-2g(x)=10g(2x)-15g(x)$
6. Figure out (not necessarily as slow as I did) that $g(nx)={n^2-n\over2}g(2x)-(n^2-2n)g(x)\tag1$ Prove by induction.
7. Plug $x\over n$ to express $g(x)$ and $g(2x)$ via $g({x\over n})$ and $g({2x\over n})$, then reverse that expression to find out that (1) is true not only for integer $n$, but for aliquot fractions as well.
8. Go from $g(x)$ to $g({x\over q})$, then to $g({px\over q})$ to prove that (1) actually holds true for any rational number in place of $n$.
9. Finally, $$g(x)=ax^2+bx\tag2$$ $$f(x,y)=a(y^2+2xy)+by\tag3$$