TITLE: Is the helix the unique path with constant curvature and constant torsion?
QUESTION [7 upvotes]: If a path has both curvature and torsion constant why it is necessarily a helix?
How to prove it?

REPLY [4 votes]: If both curvature and torsion are constant, the curve is in fact a (right) circular helix. You can, as @JyrkiLahtonen suggests, deduce this from the Fundamental Theorem of Curve Theory: By direct computation, a circular helix has constant curvature and torsion, and the curvature and torsion functions determine the curve up to rigid motion (isometry of $\Bbb R^3$).
Alternatively, with a bit of work and cleverness, you can actually explicitly integrate the Frenet equations
\begin{align*}
T'(s) &= &\kappa N(s) \\
N'(s) &= -\kappa T(s) + &&&\tau B(s) \\
B'(s) &= &-\tau N(s)
\end{align*}
to show that $\alpha(s)$ is a right circular helix. (Hint: Show that $N''(s) = -(\kappa^2+\tau^2)N(s)$.)
A generalized helix is a curve whose tangent vector makes a constant angle with a fixed vector. Such curves are characterized by the weaker condition $\tau/\kappa = \text{constant}$.