Number of intersection of two torus curves Let $(p,q)$ and $(p',q')$ be pairs of relatively prime integers. On the Torus, consider the curves $C_{p,q}$ and $C_{p',q'}$, where $C_{p,q}$ is the image of $[0,1]\ni t\mapsto (e^{i2\pi pt},^{i2\pi qt})$. Show that they intersect in exactly $|p′q−pq′|$ points.
Here, we can find similar question and a proof using homology, which is difficult to understand. Can this be proven without using homology?
 A: We have a system of equations $(e^{2\pi i pt},e^{2\pi i qt})=(e^{2\pi i p' t'},e^{2\pi i q't'})$ with $t,t'\in[0,1)$.  This is equivalent to $e^{2\pi i(pt-p't')}=1$ and $e^{2\pi i(qt-q't')}=1$, or
$$\begin{pmatrix}p&-p'\\q&-q'\end{pmatrix}\begin{pmatrix}t\\t'\end{pmatrix}\in\mathbb{Z}^2$$
The matrix has nonzero determinant $\delta=-pq'+qp'$ (relative primality of $p,q$ along with $q',p'\neq 0$ implies $\delta\neq0$).  The matrix gives a linear transformation, and we can imagine the image of $[0,1)\times[0,1)$  through it as being a parallelogram $P$, and in fact the collection of all $[a,a+1)\times[b,b+1)$ for $a,b\in\mathbb{Z}$ tesselates the plane with parallelograms, with the corners being integer points.  It follows that $\lvert\delta\rvert$, the area of $P$, is the number of integer solutions.  (Each parallelogram has the same number of integer solutions, and if we do area accounting where each parallelogram "owns" the integer unit squares whose lower-left corner lies within the parallelogram, then the area of the owned unit squares for a parallelogram must be the same as the area of the parallelogram.)
Practically speaking, you can calculate explicit solutions by finding all the integer points $(m,n)$ in $P$ (the corners of which are from the images of $(0,0),(0,1),(1,0),(1,1)$) and then using the inverse matrix to get
$$\begin{pmatrix}t\\t'\end{pmatrix}=\delta^{-1}\begin{pmatrix}-q'&p'\\-q&p\end{pmatrix}\begin{pmatrix}m\\n\end{pmatrix}$$
