 Solution

Use Y and Z as the variables appearing in the equations; the introductory reference to X and Y is a naming mismatch. With $\gamma >0$, $n>0$, a symmetric equilibrium has $Y=Z=x$ andThe left side increases continuously from zero to infinity because its derivative is $1+(n+1)x^n>0$. Thus there is exactly one positive symmetric solution for every allowed parameter pair, which verifies the implicit steady-state expression.

For $\gamma \gg 1$ and $n\gg 1$, look for one high component and one low component. If $Y\sim \gamma$, the other equation gives $Z=\gamma /(1+Y^n)\sim {\gamma }^{1-n}\ll 1$; that small value in turn gives $Y=\gamma /(1+Z^n)\sim \gamma$. Exchanging the components gives the alternative state:For example the relative correction to the high value is of order ${\gamma }^{-n(n-1)}$, while the low value first has the ordinary denominator correction of relative order ${\gamma }^{-n}$. The sharply repressive Hill function is therefore consistent with two alternatives, one factor high and the other low, which can encode a switch. Mutual repression makes a positive feedback loop: a high Y suppresses Z and thereby relieves repression of Y. Existence of alternative equilibria alone does not establish their stability; the explicit quadratic-Hill case in part (e) also permits that check.