= Solution
For a regular parameter $t$, the <curvature> and <torsion of a space curve> are
$$
\kappa=\frac{|\alpha'\times\alpha''|}{|\alpha'|^3},\qquad
\tau=\frac{\det(\alpha',\alpha'',\alpha''')}{|\alpha'\times\alpha''|^2}.
$$
The <torsion of a space curve> requires $\kappa\ne0$. Reparameterize by <arc length> $s$ and set $T=\alpha_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal <Frenet frame> shows that its <derivative> <matrix> is skew symmetric. Since $T_s=\kappa N$, defining $\tau=N_s\cdot B$ gives the <Frenet-Serret formulas>
$$
\boxed{T_s=\kappa N,\quad N_s=-\kappa T+\tau B,\quad B_s=-\tau N.}
$$
The <fundamental theorem of regular space curves> says that prescribed smooth $\kappa>0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's <linear differential equation> with an initial oriented orthonormal frame, then integrate $\alpha_s=T$; uniqueness proves the assertion.
Under the intended convention that the shared parameter $s$ is <arc length>, equality on an open interval gives equal position and equal <Frenet frames> at one point. Equal $\kappa,\tau$ give identical differential equations, so uniqueness gives $\boxed{\widetilde\alpha=\alpha\text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common <arc length>, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a <circle> into a counterexample: both curves have $\kappa=1$, $\tau=0$ and agree on $J$, but differ elsewhere as parameterized curves.
For an oriented surface with <unit normal> $n$, define <normal curvature> $\kappa_n=T_s\cdot n$ and <geodesic curvature> $\kappa_g=T_s\cdot(n\times T)$. Thus
$$
T_s=\kappa_g(n\times T)+\kappa_n n.
$$
A unit-speed curve is a <geodesic> exactly when $\kappa_g=0$. If also $\kappa_n\equiv0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a <geodesic>, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence \b[no planar patch is required, and such a <geodesic> cannot be simple and closed].
Finally let the <principal curvatures> be $\kappa_1,\kappa_2$. If the <Gaussian curvature> $K=\kappa_1\kappa_2\geq0$, they have the same sign or one vanishes. <Euler formula for normal curvature> makes $\kappa_n$ a weighted average of them. For a <geodesic>, $\kappa=|\kappa_n|$, so
$$
\boxed{\kappa\leq\max(|\kappa_1|,|\kappa_2|)\leq|\kappa_1+\kappa_2|=2|H|.}
$$
A horizontal <circle> on a cylinder of radius $R$ is a <geodesic> with $\kappa=1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.
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