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For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. 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 formulasThe 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 $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. 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 ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, 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 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 $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.

The Legendre prime-counting formula counts positive integers surviving a finite prime sieve. For $x\ge 2$, put $a=\pi (\sqrt{x})$ and let $p_1,\dots ,p_a$ be the primes at most $\sqrt{x}$. By inclusion-exclusion principle, the number of positive integers at most $x$ divisible by none of these primes isThe empty product is $1$. Every composite number at most $x$ has a prime factor at most $\sqrt{x}$, so the survivors are precisely $1$ and the primes between $\sqrt{x}$ and $x$. ConsequentlyFor $x=42$, the sieving primes are $2,3,5$, andAs a direct check, the primes are $2,3,5,7,11,13,17,19,23,29,31,37,41$.