Prove or Disprove $G(x, y)=G(y,x)$

Question

Let $G(x, y)$ be a polynomial such that: $$\frac{\partial}{\partial x}G(x, y)=\frac{\partial}{\partial y}G(x, y)$$ Prove or disprove that $G(x, y)=G(y,x)$.

Answer 1

Another approach: Let a point $(x,y)$ be given, and consider the auxiliary function $\phi(t):=G(x+t,y-t)$. The chain rule gives $$\phi'(t)={\partial G\over\partial x}\biggr|_{(x+t,y-t)}\cdot 1+{\partial G\over\partial y}\biggr|_{(x+t,y-t)}\cdot(-1)\equiv0\ ,$$ hence $\phi$ is constant. This implies $$G(x,y)=\phi(0)=\phi(y-x)=G(y,x)\ .$$ Note that this argument works for any $C^1$ function $G$ satisfying your PDE.

Answer 2

I assume that the coefficients of $G(x,y)$ are in $\mathbb{C}$ (or in some field $\mathbb{K}$ of characteristic $0$). First, we assume that $G(x,y)$ is homogeneous of degree $n$. That is, $$G(x,y)=\sum_{k=0}^n\,a_k\,x^ky^{n-k}$$ for some constants $a_0,a_1,a_2,\ldots,a_n$. Then, the condition $$\frac{\partial}{\partial x}\,G(x,y)=\frac{\partial}{\partial y}\,G(x,y)\tag{*}$$ implies that $$\begin{align}\sum_{k=0}^{n-1}\,(k+1)a_{k+1}\,x^ky^{(n-1)-k}&=\sum_{k=0}^n\,ka_k\,x^{k-1}y^{n-k}\\&=\sum_{k=0}^n\,(n-k)a_k\,x^ky^{n-k-1}=\sum_{k=0}^{n-1}\,(n-k)a_k\,x^ky^{(n-1)-k}\,.\end{align}$$ Therefore, $$(k+1)a_{k+1}=(n-k)a_k$$ for all $k=0,1,2,\ldots,n-1$. By induction, we have $$a_k=\binom{n}{k}\,a_0\text{ for }k=0,1,2,\ldots,n\,.$$ That is, $$G(x,y)=a_0\,(x+y)^n\,.$$ From this observation, it follows that any (possibly non-homogeneous) $G(x,y)\in\mathbb{C}[x,y]$ that satisfies (*) is of the form $G(x,y)=f(x+y)$ for some $f(t)\in\mathbb{C}[t]$. The claim follows immediately.

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The claim is false in positive characteristics. Suppose that $p$ is a prime natural number and the base field $\mathbb{K}$ has characteristic $p$. Then, $G(x,y)=x^p$ satisfies (*), but $G(x,y)\neq G(y,x)$.