If a curve has velocity and acceleration of constant magnitude, then its curvature is constant. If a curve has velocity and acceleration of constant magnitude, then its curvature is constant.
I think it's too easy a problem, but I'm not sure my solution is correct:
Let $\alpha (t) = (x (t), y (t)) $ be the curve, having constant velocity and acceleration implies that $ x' (t) = c_1 $, $ y' (t) = c_2 $, $ x'' (t) = c_3 $, $ y'' (t) = c_4 $.
Using the following equation for curvature in which the parametrization does not matter and assuming that the curve is regular:
$$
\kappa=\dfrac{x'y''-x''y'}{(x'^2+y'^2)^{\frac{3}{2}}}=\dfrac{c_1c_4-c_3c_2}{(c_1^2+c_2^2)^{\frac{3}{2}}}.
$$
So with this we conclude that the curvature is constant since the denominator is different from 0.
Is the proof correct?

Hint 1. Show that the angle between $\alpha''$ and $J(\alpha')= (-y',x')$ is constant. Subhint: For this, show that the angle between $\alpha''$ and $\alpha'$ is constant.
Hint 2. Note that
$$
\kappa = \frac{\alpha'' \cdot J\alpha'}{\|\alpha'\|^3}.
$$
