O'Neill's Elementary Differential Geometry, problem 4.3.11, page 156, (Kindle edition), asks the student to discuss the periodicity of the curve $x(at,bt)$ where x is the parametrization of the torus:
$$x(u,v)=((R+r \cos(u))\cos(v),(R+r \cos(u))\sin(v),r\sin(u))$$
And $a/b$ is a rational number $m/n$. I got as far as convincing myself that $2\pi m/a=2\pi n/b$ is a period, where $m$ and $n$ have no common divisor.
$$x(t)=((R+r\cos(at)\cos(bt),(R+r\cos(at))\sin(bt),r\sin(at))$$
Any period of the curve is an integer multiple of the least period of the third component, $2\pi/a$. Since $R>r$, $R+r\cos(at)>0$ and any period is also an integer multiple of $2\pi/b$. The ratio of these two integers is $a/b=m/n$, so the two integers are $m$ and $n$ for the least period. Does this sound correct?
Problem goes on to ask the student to show that the curve $x$ has no self-crossings. According to the first edition of the book, this means that form $x(t)=x(t+T)$ follows that $T$ is a multiple of the least period. How to show that this is the case?
Problem goes on to ask the student to show that if $a/b$ is irrational then $x(at,bt)$ is one-to-one. I guess the thing to do is to show that if it is not one-to-one then $a/b$ is rational. I have no idea how to do this and would like a hint.
This last case is interesting because $x$ comes arbitrarily close to every point on the torus.
Best Answer
Your reasoning about period of curve seems correct.
To answer second question, about self-crossing, you should again inspect components of curve, and with similar reasoning deduce that $T$ is multiple of period.
Third part of problem now follows from first two. If $x$ is not one-to-one, than there is some point of self intersection (e.g for some $t$ and $T$ holds $x(t)=x(t+T)$). From second part we conclude that $T$ is multiple of period. Now from first part follows that period is rational.