Number of fixed points of a hyperbolic toral automorphism Any integer matrix $A\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ induces a map $T_{A}\colon\mathbb{R}^{2}/\mathbb{Z}^{2}\to\mathbb{R}^{2}/\mathbb{Z}^{2}$ defined by $$T_{A}((x,y)+\mathbb{Z}^{2}):=A(x,y)+\mathbb{Z}^{2}$$ for all $(x,y)\in\mathbb{R}^{2}$. The map $T_{A}$ is called a toral automorphism if and only if $A^{-1}$ exists and is an integer matrix (or, equivalently: $|\text{Det}(A)|=1$). The map $T_{A}$ is called hyperbolic if and only if the eigenvalues of $A$ do not lie on the unit circle in $\mathbb{C}$. A point $(x,y)+\mathbb{Z}^{2}$ is called a fixed point of  $T_{A}$ if and only if $$T_{A}((x,y)+\mathbb{Z}^{2})=(x,y)+\mathbb{Z}^{2}.$$
Let $\text{Fix}(T_{A})\subset\mathbb{R}^{2}/\mathbb{Z}^{2}$ denote the set consisting of all the fixed points of $T_{A}$. I want to prove that, if $T_{A}$ is a hyperbolic toral automorphism, then $$|\text{Fix}(T_{A})|=\text{Det}(A-I),$$ where $I\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ is the identity matrix.
MY ATTEMPT: Let $x,y\in[0,1)$ and observe that $(x,y)+\mathbb{Z}^{2}$ is a fixed point of $T_{A}$ if and only if there exist $m,n\in\mathbb{Z}$ such that $A(x,y)=(x,y)+(m,n)$, which is equivalent to $$(A-I)(x,y)=(m,n).$$ Hence, it follows that $$|\text{Fix}(T_{A})|=|\{(A-I)(x,y) \ | \ x,y\in[0,1)\}\cap\mathbb{Z}^{2}|.$$ Let $P\subset\mathbb{R}^{2}$ be the parallelogram spanned by the vectors $(A-I)(1,0)$ and $(A-I)(0,1)$. Then $\text{Area}(P)=\text{Det}(A-I)$. Note that $A$ maps the unit square onto $P$. Hence, it remains to show that the number of integer pairs $(m,n)$ in $P$ is precisely the area of $P$.
MY QUESTIONS: 
1) How do I proceed?
2) Do I really need that $T_{A}$ is a hyperbolic automorphism?
 A: Your question leads to the following fact: If $L \subset \mathbb R^2$ is a lattice spanned by integer vectors, then the number of integer points in $\mathbb R^2 / L$ is the covolume of $L$ (area of a fundamental parallellogram for $L$).
This may be deduced either from


*

*Pick's formula: Call $i$ the number of interior lattice points of $P$ and $b$ the number of points on the boundary. Call $x$ the number of points on one side and $y$ the number on an adjacent side. Then $b = 2x+2y-4$ (we counted the four corners twice). Then the number of lattice points in $\mathbb R^2 / L$ is $i + x + y - 3$: By taking $x+y$, we counted the point in the corner four times, whereas we only want it once. By Pick's formula, the volume of $P$ is $i + b/2 - 1$. The two are equal: $i+b/2 - 1 = i + (2x+2y-4)/2-1 = i + x + y - 3$.

*The following intuitive argument (which can be made hard). Let $P$ be a fundamental parallellogram for $L$ and $n$ the number of integer points in $\mathbb R^2 / L$. Tile together $N^2$ copies of $P$ to obtain the parallellogram $NP$ for some large $N \in \mathbb N$. The volume of $NP$ is $N^2 \cdot \mathrm{Vol}(P)$, so the number of integer points in this region is $N^2 \cdot \mathrm{Vol}(P) + O(N)$. On the other hand, the number of integer points is $N^2 \cdot n + O(N)$. Taking $N\to \infty$, we see that we must have $n = \mathrm{Vol}(P)$.

*The following argument: Let $B$ be a matrix that sends $\mathbb Z^2$ to $L$. Then $B$ gives a linear map $\mathbb Z^2 \to \mathbb Z^2$ and we want to count the index $[\mathbb Z^2 : B(\mathbb Z^2) ]$. From the Smith normal form for $B$, we see that this index equals $|\det(B)|$.


So you don't need the matrix $A$ to be hyperbolic, you also don't need $|\det A| = 1$; all you need is that $A - I$ is invertible, which is the case if $A$ is hyperbolic.
A: Let me do the case that $\text{det}(A)=1$, equivalently $T_A$ preserves orientation. In that case, your formula quickly converts into $|\text{Fix}(T_A)| = \text{trace}(A)-2$. 
One standard proof of this formula uses Markov partitions. Here's a very brief outline.
Step 1 is to conjugate $A$ to be a positive matrix, which can always be done for hyperbolic total automorphisms that preserve orientation. 
Step 2 is to construct a Markov partition with $2$ Markov boxes $B_1,B_2$, whose transition matrix is equal to $A$; for example here is a depiction of the Markov partition for $\begin{pmatrix}2 & 1 \\ 1 & 1 \end{pmatrix}$. To say that the transition matrix is equal to $A$ means that $A(i,j)$ is equal to the number of components of the open set $\text{int}(B_i) \cap T_A(\text{int}(B_j))$; let me refer to these as the "$(i,j)$ overlap components".
Step 3 is to look at each $(i,i)$ overlap components, show that its closure contains a unique fixed point of $T_A$, and show that every fixed point arises in this manner. 
Step 4 is to notice that there is a slight overcounting of fixed points: a fixed point occurring on the boundary of an overlap component may occur on a boundary of another overlap component. But this over counting can be precisely counted, and one only has to subtract $2$.
Thus, the formula $\text{trace}(A)-2$. 
As a final comment, yes, you need $T_A$ to be a hyperbolic automorphism. You can isotope the automorphism $T_A$ to produce homeomorphisms of the torus with an arbitrarily high number of fixed points, in fact you can isotope it to produce an entire open subset of fixed points.
