Let $\{e, f\}$ be an orthonormal basis of $E^2$

Question

Let $\{e, f\}$ be an orthonormal basis of $E^2$. Consider a linear operator P such that $a=P(e)=91e+50f$ and $b=P(f)=20e+11f$ Consider the parallelogram $\Pi_{a,b}$ formed by $a$ and $b$ attached to the origin. Find the area. Show that the vertices of $\Pi_{a,b}$ are the only points of $\Pi_{a,b}$ whose coordinated are integers. $$$$ I can find the area and had a go at the next problem using the fact that any for any vector $k$ on the parallelogram either $(k-a)||b$ or $(k-b)||a$ and then using diophantine eqs. But that approach didn't work

Answer 1

The area of $\Pi_{a,b}$ is simply the determinant of the matrix with the vectors $a,b$ as columns:

$$\operatorname{Area}(\Pi_{a,b}) = \begin{vmatrix} 91 & 20 \\ 50 & 11\end{vmatrix} = 1$$

The set $\Pi_{a,b}$ can be expressed as $$\Pi_{a,b} = \{\alpha a + \beta b : \alpha,\beta \in [0,1]\}$$

If $\alpha a + \beta b$ has integer coordinates, then there are $m,n \in \mathbb{Z}$ such that $$(91\alpha + 20\beta)e + (50\alpha+11\beta)f = \alpha a + \beta b = me + nf$$

or in matrix form $$\begin{bmatrix} 91 & 20 \\ 50 & 11\end{bmatrix}\begin{bmatrix}\alpha \\ \beta\end{bmatrix} = \begin{bmatrix}m\\ n\end{bmatrix} \implies \begin{bmatrix}\alpha \\ \beta\end{bmatrix} = \begin{bmatrix} 91 & 20 \\ 50 & 11\end{bmatrix}^{-1}\begin{bmatrix}m\\ n\end{bmatrix} = \begin{bmatrix}11m-20n\\ -50m+91n\end{bmatrix}$$

so $\alpha = 11m-20n \in \mathbb{Z}$ and $\beta = -50m+91n\in\mathbb{Z}$. Since the only integers in $[0,1]$ are $0$ and $1$, we conclude that $\alpha,\beta \in \{0,1\}$.

Therefore, the only points in $\Pi_{a,b}$ with integer coordinates are $0,a,b$ and $a+b$ which are precisely the vertices.