= Lozi fixed points and the alternating-sign two-cycle
{c}
{title2=$x_1=(a+1-b)/(a^2+(1-b)^2),\quad x_2=(1-b-a)/(a^2+(1-b)^2)$}
The positive <fixed point> exists when $b<1+a$ and equals $(1,b)/(1+a-b)$; the negative one exists when $b>1-a$ and equals $(1,b)/(1-a-b)$. Equality on either boundary does not give a branch, because its fixed-point equation becomes $0=1$. For a genuine <two-cycle> and $a^2\ne(1-b)^2$, subtracting its scalar equations rules out equal signs of the two first coordinates. Solving with $x_1>0>x_2$ gives the displayed coordinates and second coordinates $y_1=bx_2$, $y_2=bx_1$. Their signs require exactly $a>|1-b|$. The cycle is unique up to interchanging its two points. Degenerate boundary <parameter> lines can support additional period-two families and are outside this statement.
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