= Solution
For a repulsive short-range potential and a stable finite minimum, assume $A>0$ and integer $n>1$. Setting $u'(R)=C/R^2-2nA/R^{n+1}=0$ gives
$$
\boxed{R_0=\left(\frac{2nA}{C}\right)^{1/(n-1)}
=\left(\frac{8\pi\epsilon_0nA}{\alpha q^2}\right)^{1/(n-1)}.}
$$
At this point $2A/R_0^n=C/(nR_0)$, so
$$
\boxed{u_0=-\frac{C}{R_0}\left(1-\frac1n\right),\qquad
u''(R_0)=\frac{(n-1)C}{R_0^3}>0.}
$$
Here $u''$ denotes the second derivative of the <energy> $u$, not a displacement variable. The negative equilibrium <energy> is the cohesive <energy> per pair relative to infinite separation.
The qualifiers on $A,n$ matter: an arbitrary real coefficient and integer exponent do not automatically describe stabilizing repulsion. For instance $A<0,n>1$ gives collapse as $R\to0$, whereas $n=1$ gives $u=(2A-C)/R$ and has no isolated finite equilibrium. The displayed minimum therefore applies to the physically repulsive, faster-than-Coulomb branch of the stated model.
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