= Solution
The <proper acceleration> is $\alpha=\sqrt{-A^2}$, the magnitude of the instantaneous rest-frame three-acceleration. A <pure force> is a <four-force> preserving <rest mass>, or equivalently satisfying $F\cdot U=0$. Indeed $F=d(mU)/d\tau$ gives $F\cdot U=c^2dm/d\tau$.
For the right-hand branch of the hyperbola, use $c=1$, $x>0$ and $x=\sqrt{L^2+t^2}$, taking $L>0$. Direct differentiation gives
$$
v=\frac{dx}{dt}=\frac t x,\qquad
|v|=\sqrt{1-\frac{L^2}{x^2}},\qquad
\gamma=\frac xL,\qquad
a=\frac{dv}{dt}=\frac{L^2}{x^3}.
$$
The sign of $v$ follows the sign of $t$, whereas the coordinate acceleration on this branch is positive. Consequently
$$
\boxed{a\gamma^3=\frac1L=\alpha.}
$$
The proper-time parameterization $t=L\sinh(\tau/L)$, $x=L\cosh(\tau/L)$ independently gives constant <proper acceleration>. Restoring dimensions, a hyperbola $x^2-c^2t^2=L^2$ has $\alpha=c^2/L$. Constant proper acceleration does not mean that all inertial-frame components of the <four-force> are constant.
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