= Solution
There is no nonzero, literally constant <four-force> that is a <pure force> throughout this motion. Initially $U=(c,0,0,0)$, so purity forces $F^0=0$. If the full four-vector were constant, $U(\tau)=U(0)+F\tau/M_0$; its square would be $c^2+F^2\tau^2/M_0^2$, which cannot remain $c^2$ for a nonzero initially spatial $F$. This is a genuine ambiguity in the source wording.
The usual consistent interpretation is constant laboratory <relativistic force> $f=dp_x/dt$, equivalently constant one-dimensional <proper acceleration> $f/M_0$. Take $f>0$ and let $s=ft/(M_0c)$. Integrating <momentum conservation> with the applied impulse gives $p_x=ft$, so
$$
\boxed{U^\mu(t)=c\left(\sqrt{1+s^2},s,0,0\right),\qquad
\beta(t)=\frac{s}{\sqrt{1+s^2}},\qquad\gamma(t)=\sqrt{1+s^2}.}
$$
The corresponding <four-force> has components $F^\mu=\gamma f(\beta,1,0,0)$, which are not constant, although its invariant magnitude is $F^2=-f^2$. Its temporal component is indeed $F^0=(\gamma/c)dE/dt$. The speed rises from zero and approaches $c$ asymptotically; $\gamma$ starts at one, initially rises quadratically and becomes linear in $t$ at late times.
If instead only the spatial <four-force> component $F^x=f_x$ is held constant, the distinct consistent answer is most simply parametrized by $u=f_x\tau/(M_0c)$:
$$
\boxed{U^\mu=c(\sqrt{1+u^2},u,0,0),\qquad
\frac{f_xt}{M_0c}=\frac12\left[u\sqrt{1+u^2}+\operatorname{arsinh}u\right],\qquad
F^0=\frac{f_xu}{\sqrt{1+u^2}}.}
$$
This follows by integrating $dU^x/d\tau=f_x/M_0$ and $dt/d\tau=U^0/c$. The temporal component must still vary. These two interpretations must not be conflated with a constant four-vector.
\Image[/past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2017-constant-force-interpretations.png]
{title=Velocity and Lorentz factor for two consistent constant-force interpretations}
{description=Solid curves hold laboratory three-force constant; dashed curves hold only the spatial component of four-force constant. Both use the same force scale and start from rest. A nonzero constant complete four-force cannot preserve rest mass throughout the motion.}
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