Proving that a set is infinite Consider $G\subset M_2(\mathbb{C})$ where $G =\begin{Bmatrix}
\begin{pmatrix}
a & 10b\\ 
b & a
\end{pmatrix} 
& | a,b \in \mathbb{Q},a^2-10b^2=1& 
\end{Bmatrix}
$ . Prove that G is infite.
I was thinking of searching for two numbers $a$ and $b$ from $\mathbb{Q}$ who don't satisfy the equation $a^2-10b^2=1$ , but how?
 A: You need to prove that there are infinitely many $a,b$ such that 
$$a^2-10b^2=1$$
It is easy to find one, and then if you look for Pell-Fermat equation( aka Pell equation) you will see how to construct infinitely many solutions.
A: Note that if $A,B\in G$ then $AB\in G$ hence if :
$$ A=\begin{pmatrix}19 &60 \\ 6 &19\end{pmatrix}\in G$$
Now prove that $A^n\in G$ for every positive integer $n$



*

*Given two elements of $G$ :
$$A=\begin{pmatrix}a &10b \\ b&a\end{pmatrix} \,\, B=\begin{pmatrix}a&10b' \\ b' &a' \end{pmatrix} \implies AB=\begin{pmatrix}aa'+10bb' &10(ab'+ba')\\ ab'+ba'&aa'+10bb'\end{pmatrix}\in G$$
hence $G$ is closed by multiplication and for every $A,B\in G$ we have $AB\in G$ and in particular $tr(AB)\geq 2aa'=\frac{tr(A)tr(B)}{2}$

*Now it's easy to observe that:
$$ M=\begin{pmatrix}19 &60 \\ 6 &19\end{pmatrix}\in G$$
and as a consequence $M^n\in G$ for evry positive integer $n$, and we know that $$tr(M^{n+1})\geq \frac{tr(M^n)tr(M)}{2}=19tr(M^n)>tr(M^n)$$
so the sequence $M^n$ is an infinte sequence of elements of $G$

