Relationship between trace of a linear map and the number of points it fixes. Problem Statement: Let $\Phi_A:T^2\rightarrow T^2$ be a smooth mapping into the torus induced by a linear map $A\in SL_2(\mathbb{Z})$ under the quotient relation that identifies 0 and 1. Assume that A does not have 1 as an eigenvalue. Show that the number of fixed points can be expressed in terms of the Trace of A only.
Thoughts: The condition that $\Phi_A$ has a fixed point is equivalent to $A$ taking a point $(a,b)\in \mathbb{R}^2$ to $(a+m,b+n)$ where $m,n\in \mathbb{Z}$. Multiplying the matrix by a vector $(a,b)$ with entries $\alpha, \beta, \gamma, \delta$ we obtain equations $(a,b)=(\alpha a + \beta b, \gamma a + \delta b)$. Those can only differ by the addition of an integer, so $\beta = \gamma = 0$ and then the only two elements of the matrix left are the eigenvalues - the terms which are found in the Trace. and if A takes something to ($\alpha a, \beta b$) then $\Phi_A$ must take it to $(a,b)$ since they differ by an integer. 
I can't seem to get the equation for the # of fixed points out of this list of observations. Help appreciated, will upvote. 
EDIT: Solution
If we substitute the A into its characteristic polynomial of A and use Cayley Hamilton, we get: $A^2(a,b)+tr(A)A(a,b)+1=0$ which is equivalent to $(a,b)+A(m,n)−tr(A)(a,b)−tr(A)(m,n)+(a,b)=0$ which we can solve for $(a,b)$ to get... $2−tr(A)(a,b)=(A+I−tr(A))(m,n)$. This line tells us that that $2−tr(A)(a,b)∈\mathbb{Z}$, i.e, $(a,b)$ are multiples of $12−tr(A)$ generated by some $(m,n).$ Furthermore, we have $(a,b)=12−tr(A)(A+I−tr(A)(m,n).$ The linear map on the right side has $detA=2−tr(A).$ 
That means that the unit parallelogram of the lattice has area $2−tr(A)$ and therefore in each unit square, there will be $2−tr(A)$ copies of the unit parallelogram lying inside the unit square. The lattice points lying inside the unit square are precisely those which don't differ by some $(m,n)∈\mathbb{Z}$. Since there are $2−tr(A)$ copies in the Unit square, we'll have 2−tr(A) corner points, and those points each correspond to an equivalence class of solutions.
 A: Yes, interpreting correctly, the Lefschetz number $\lambda_f$ of the map counts the fixed points. This is — in the differential topology world — the intersection number of the diagonal in $X\times X$ with the graph of $f$. This can be computed in terms of the algebraic topology: 
$$\lambda_f = \sum (-1)^k \text{tr}\big(f_*\colon H_k(X) \to H_k(X)\big).$$
In the case of your problem, this is going to be $2-\text{tr}A$.
A: Try writing the equations as $A \begin{pmatrix}a \\ b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix}m \\ n \end{pmatrix}.$ (Here $(a,b)$ is a point in $\mathbb R^2$, and $(m,n)$ is a point in $\mathbb Z^2$.)  
Rewriting, this gives $(A - I)\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix}m \\ n \end{pmatrix}.$
Now the number of fixed points is the number of inequivalent solutions to this equation, where two solutions are regarded as equivalent if their $(a,b)$ differ by something in $\mathbb Z^2$.  
You can just solve this equation by linear algebra, if you just choose some $(m,n)$, but note that while $A- I$ has integer entries, its inverse may not --- and if you compute $\det A - I$, you will see the role of the trace of $A$.  So choosing different $(m,n)$ can give non-equivalent solutions, and if you think carefully about how many equivalence classes of solutions you can get, you will find your desired formula.

Added at the OP's request:  First we compute that $\det( A - I ) = 2 - \mathrm{tr} A.$  By assumption this is non-zero.  
Now we have the following problem: let $M$ be a $2 \times 2$ matrix with integer entries and non-zero determinant; compute the number of solutions
to $M \begin{pmatrix} a \\ b \end{pmatrix} \in \mathbb Z^2,$ where the
solutions are also counted modulo $\mathbb Z^2$.   The answer is: the number of solutions is equal to $| \det M \, |$.  
One way to see this is to use the theory of elementary divisors to reduce to the case when $M$ is diagonal.   Another way, which should at least make intuitive sense, is to see that the preimage of a unit square under multiplication by $M$ will have area $| \det M \, |^{-1}$, so we will be able to find $| \det M \, |$ such preimages which don't differ by an element of $\mathbb Z^2$.
