# Cauchy-like functional equation

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.

• It seems like plugging $x = y = 0$ in gives you $f(z,0) = - f(0,0)$ for all $z$, and so, by having $z = 0$, we get $f(0,0) = -f(0,0) = 0$, and as such $f(z,0) = 0$ for all $z$. Then, by plugging $y = 0$, we have: $f(0,z) + f(z,x) = f(0,x+z)$, but I don't know if that's useful yet. Commented Feb 24, 2023 at 9:32

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, $$g(x)=ax^2+bx\tag2$$ $$f(x,y)=a(y^2+2xy)+by\tag3$$
• I struggle with understanding steps 7 and 8. Doing the plug I get $g(x) = \frac{n^2-n}{2} g(\frac{2x}{n}) - (n^2-2n)g(\frac{x}{n})$ and I don't know how to deal with the $g(\frac{2x}{n})$ term. Am I supposed to arrive at a result like $g( \frac{x}{n} ) = \frac{(\frac{1}{n})^2-\frac{1}{n}}{2}g(2x)-((\frac{1}{n})^2-2\frac{1}{n})g(x)$? Commented Mar 3, 2023 at 10:09
• Yes, like that. To get another equation, plug $x\over n$ instead of $x$ and $2n$ instead of $n$. That would give you $g(2x)$ in terms of $g(x/n)$ and $g(2x/n)$. Commented Mar 3, 2023 at 12:49
• I am still lost. What would the substituted equation look like? And I suppose it takes a lot of algebraic manipulation to get to the desired result? Do I need to rearrange to put $g(\frac{x}{n})$ first? Like $g(\frac{x}{n})=\frac{1}{2n^2-n} (g(x) + (4n^2-4n)g(\frac{x}{2n}))$ Commented Mar 3, 2023 at 16:43
• The substituted equation would be $g(2x) = \frac{(2n)^2-2n}{2} g(\frac{2x}{n}) - ((2n)^2-2\cdot2n)g(\frac{x}{n})$. The rest is no more advanced than something like $$\begin{cases}5a+3b&=11\\7a-2b&=3,\\ \end{cases}$$ except that you have some algebraic expressions in place of "5", "3", "11", etc. Also, you already know the end result. Commented Mar 3, 2023 at 17:58