= 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 $Y=Z=x$ and
$$
\boxed{x(1+x^n)=\gamma.}
$$
The 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 $\gamma\gg1$ and $n\gg1$, look for one high component and one low component. If $Y\sim\gamma$, the other equation gives $Z=\gamma/(1+Y^n)\sim\gamma^{1-n}\ll1$; that small value in turn gives $Y=\gamma/(1+Z^n)\sim\gamma$. Exchanging the components gives the alternative state:
$$
\boxed{(Y,Z)\sim(\gamma,\gamma^{1-n})
\quad\text{and}\quad(\gamma^{1-n},\gamma).}
$$
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.
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