= Cubic swap-map two-cycle factorization
{title2=$(x+y)(axy/3-1)=0$}
For $F(x,y)=(y,a(x-x^3/3)-y)$, a genuine <two-cycle> obeys $F(x,y)=(y,x)$ with $x\ne y$. Subtracting its two component equations gives $x^2+xy+y^2=3$, then substitution gives the displayed factorization. Cancelling $x+y$ loses the antipodal cycle $(\sqrt3,-\sqrt3)\leftrightarrow(-\sqrt3,\sqrt3)$, present for every $a>0$. The nonzero-sum family instead has $xy=3/a$ and $x+y=\pm\sqrt{3+3/a}$, requiring $a>3$. Retaining all vanishing factors is essential in <polynomial> cycle classification.
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