= Solution
With $\sigma=1$, substitute $x=(u+v)/2$, $y=(u-v)/2$, $r=1+\mu$. The equations become
$$
\boxed{\dot u=\tfrac12(\mu-z)(u+v),\quad
\dot v=-2v-\tfrac12(\mu-z)(u+v),\quad
\dot z=\tfrac14(u^2-v^2)-bz,\quad\dot\mu=0.}
$$
The extended centre variables are $u,\mu$. The <centre manifold theorem> gives smooth graph functions $V,Z$ with $V(0,0)=Z(0,0)=0$ and all first derivatives zero. Their invariance equations are $V_u\dot u=\dot v$ and $Z_u\dot u=\dot z$, after substitution of the graph.
Count $\mu$ as order $u^2$. Then $\dot u=O(u^3)$ on the graph, so the left side of the $Z$ equation contributes only at higher order. Its quadratic terms give $0=u^2/4-bZ$, hence $Z=u^2/(4b)+O(u^3)$. The cubic terms of the $V$ equation similarly give $0=-2V-\mu u/2+uZ/2$. Thus
$$
\boxed{Z=\frac{u^2}{4b}+O(u^3),\qquad
V=-\frac{\mu u}{4}+\frac{u^3}{16b}+O(u^4).}
$$
Terms purely in $\mu$ vanish because the origin remains an <equilibrium> for every parameter; these expansions describe the weighted scaling $\mu=O(u^2)$, not an unrestricted one-variable Taylor <series>. Substitution gives the reduced equation
$$
\boxed{\dot u=\frac\mu2u-\frac1{8b}u^3+O(u^4).}
$$
Its cubic coefficient is negative. For $\mu<0$ the origin is attracting in the centre direction; for $\mu>0$ it loses stability and two attracting branches emerge with $u\sim\pm2\sqrt{b\mu}$. Therefore the bifurcation is a \b[supercritical pitchfork]. These branches agree with the exact Lorenz equilibria $x=y=\pm\sqrt{b\mu}$, $z=\mu$.
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