Can an $n$-armed ($n\in\mathbb N$) spiral in $\mathbb R^2$ be characterised by the equation $F(x)+G(y)=0$, where $F$ and $G$ are some continuous real-valued functions.
Definition: An $n$-armed spiral is a union of $n$ curves $C_i\subset \mathbb R^2, i=1,\ldots,n$ that other than at the origin don't intersect and each $C_i$ can be represented in polar coordinates as $$ C_i = \Big\{ \Big( r_i(t) \sin\big(\phi_i(t)\big), r_i(t) \cos\big(\phi_i(t)\big) \Big), t\in \mathbb R_+ \Big\}, $$ where both $r_i$ and $\phi_i$ are continuous increasing functions such that $r_i(0)=0$, and $\phi_i(t)\to\infty$ as $t\to\infty$.
(The assumption that both $r_i$ and $\phi_i$ are increasing can be replaced by assuming that the curve $C_i$ does not intersect itself.)
I guess that if there is such an $n$-armed spiral, then it will be symmetric so that its individual arms can be obtained from each other by rotation by $2\pi/n$ about the origin.
@HagenvonEitzen showed in a very elegant way that for $n=1$ the answer is no. The argument was based on the fact that $\mathbb R^2 \setminus C_1$ forms a single connected set on which the function $F(x)+G(y)$ can not change sign. In the case of $n$ arms, $\mathbb R^2 \setminus (C_1\cup \ldots \cup C_n)$ consists of $n$ connected sets / regions. If $n$ is odd, then at least on two of the neighbouring regions the function $F(x)+G(y)$ must have the same sign, and so the @HagenvonEitzen's argument applies and it can be concluded than an even-number armed spiral cannot be represented in the desired way. What can we tell for $n$ even?