= Solution
For <integers> $k,l$,
$$
A\begin{pmatrix}x+k\\y+l\end{pmatrix}-A\begin{pmatrix}x\\y\end{pmatrix}
=\begin{pmatrix}2k+l\\k+l\end{pmatrix}\in\mathbb Z^2.
$$
Thus reducing coordinates <modulo one> gives a well-defined <function> on the <torus> $\mathbb R^2/\mathbb Z^2$. Its <determinant> is one and its inverse <matrix> is the <integer matrix> $\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$, so it induces a <bijection> of the <torus>. The <Jacobian determinant> one makes this <bijection> <area-preserving>.
Its <eigenvalues> solve $\lambda^2-3\lambda+1=0$, giving
$$
\boxed{\lambda_+=\frac{3+\sqrt5}{2}>1,\qquad\lambda_-=\frac{3-\sqrt5}{2}=\lambda_+^{-1}<1.}
$$
There are therefore expanding and contracting <eigendirections> and no <eigenvalue> on the <unit circle>: the <function> is a <hyperbolic toral automorphism>. The <Lyapunov exponents> per iterate are
$$
\boxed{\chi_+=\log\frac{3+\sqrt5}{2},\qquad\chi_-=-\log\frac{3+\sqrt5}{2}.}
$$
They sum to zero, as required by <area preservation>.
The four vertices of the unit <square> map to $(0,0),(2,1),(3,2),(1,1)$, the vertices of an area-one <parallelogram>. To reduce it <modulo one>, cut it by the <integer> grid. For each piece in the cell $[j,j+1]\times[k,k+1]$, translate by $(-j,-k)$. Up to boundary identifications these translated pieces partition the unit <square>, because the inverse <integer matrix> gives exactly one <torus> preimage to every point. The colored pieces in the original diagram show the same pieces before and after these translations.
\Image[/past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2008-cat-map.png]
{title=Area-preserving torus map by cutting and translating its parallelogram}
For the <Fibonacci> identity, at $n=1$ the claimed <matrix> is $\begin{pmatrix}F_2&F_1\\F_1&F_0\end{pmatrix}=A$. If the identity holds at $n$, multiplication on the right by $A$ gives
$$
A^{n+1}=\begin{pmatrix}2F_{2n}+F_{2n-1}&F_{2n}+F_{2n-1}\\2F_{2n-1}+F_{2n-2}&F_{2n-1}+F_{2n-2}\end{pmatrix}
=\begin{pmatrix}F_{2n+2}&F_{2n+1}\\F_{2n+1}&F_{2n}\end{pmatrix}.
$$
Each equality uses $F_{j+1}=F_j+F_{j-1}$, with the specified indexing $F_0=F_1=1$. This proves the formula for every positive <integer> $n$.
A <fixed point> satisfies $(A-I)v\in\mathbb Z^2$. Since $A-I=\begin{pmatrix}1&1\\1&0\end{pmatrix}$ has an inverse <integer matrix>, this forces $v\in\mathbb Z^2$, so \b[the origin is the only <fixed point> <modulo one>].
For period dividing two,
$$
(A^2-I)v=\begin{pmatrix}4&3\\3&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\in\mathbb Z^2.
$$
The second congruence gives $y=-3x\pmod1$; inserting it into the first gives $5x=0\pmod1$. Thus the complete list is
$$
(x,y)=\left(\frac j5,-\frac{3j}5\bmod1\right),\qquad j=0,1,2,3,4.
$$
The first is the fixed origin. Direct application of $A$ exchanges the other four in exactly two pairs:
$$
\boxed{\{(1/5,2/5),(4/5,3/5)\},\qquad\{(3/5,1/5),(2/5,4/5)\}.}
$$
This both verifies the <periodic orbits> and excludes additional ones.
Finally, if $v$ is fixed by $F^p$, then $(A^p-I)v=m$ for an <integer> vector $m$. The <eigenvalues> of $A^p-I$ are $\lambda_+^p-1$ and $\lambda_-^p-1$, neither zero. Hence its inverse exists and has <rational> entries by the <adjugate matrix> formula for an <integer matrix>. Therefore $v=(A^p-I)^{-1}m$ has <rational> coordinates. The <derivative> of the <return map> is $A^p$, with <eigenvalues> $\lambda_+^p>1$ and $0<\lambda_-^p<1$. \b[Every <periodic point> is a hyperbolic <saddle fixed point of a map> and is <unstable>], although <perturbations> in its <stable> <eigendirection> contract.
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