= Solution
The <quadratic-Hill mutual repression switch> is explicit: for $n=2$, <steady state> requires $Y(1+Z^2)=\gamma$ and $Z(1+Y^2)=\gamma$. Subtraction gives
$$
(Y-Z)(1-YZ)=0.
$$
Thus either $Y=Z=x$, giving the unique positive cubic root $x^3+x-\gamma=0$, or $YZ=1$. In the latter case, $Y+1/Y=\gamma$, hence
$$
\boxed{Y_\pm=\frac{\gamma\pm\sqrt{\gamma^2-4}}2,\qquad
Z_\pm=\frac1{Y_\pm}=Y_\mp.}
$$
The non-symmetric pair is real, positive and distinct exactly when $\gamma>2$. For $0<\gamma<2$ only the symmetric solution exists. Eliminating one variable gives the source's hinted factorization,
$$
(Y^3+Y-\gamma)(Y^2-\gamma Y+1)=0,
$$
with no singular denominator for real <concentrations>. The cubic is strictly increasing since $3Y^2+1>0$, so it contributes only the symmetric state.
One can verify the switch interpretation without a lengthy <stability> analysis. At symmetry the <Jacobian matrix> <eigenvalues> are $-1\pm2x^2/(1+x^2)$: symmetry is stable for $x<1$ and becomes a saddle for $x>1$, corresponding to $\gamma<2$ and $\gamma>2$. At either asymmetric <equilibrium> the <eigenvalues> are $-1\pm2/\gamma$, both negative when $\gamma>2$. Therefore
$$
\boxed{\gamma>2:\ \text{two attracting alternatives and a separating saddle};\qquad
0<\gamma<2:\ \text{one attracting symmetric state}.}
$$
At $\gamma=2$, all three branches meet at $(1,1)$ with a zero linear <eigenvalue>, the switching threshold rather than two distinct robust memory states. The lower panels of the original figure in <past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2018/4/a/solution> show representative <nullclines> on either side.
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