I would like to determine the elements of order two in $GL_2(\mathbb{Z})$. I reduced the problem to solve the diophantine equation $a^2+bc=1$ with $a,b,c \in \mathbb{Z}$, but I have no idea to solve it.
Elements of order two in $GL_2(\mathbb{Z})$
1 Answers
How did you arrive to only that condition?? You also need other ones:
$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\,\,,\,\,A^2=I\Longrightarrow bc=0\,\,or\,\,tr.(A)=a+d=0\,\,\,and\,\,\,a^2+bc=d^2+bc=1$
$\text{and also}\,\,A^2=1\Longrightarrow (\det A)^2=1\Longrightarrow ad-bc=\pm 1$
Now, cases:
$(1)\;\;\;b=0\;\Longrightarrow a^2=d^2=1\Longrightarrow a,d=\pm 1\,\,,\,b=0\,\,,\,c\in\Bbb Z$
$(2)\;\;\;c=0 --\text{just as above, mutatis mutandis} --$
$(3)\;\;\;bc\neq0\;\Longrightarrow a=-d\,\,,\,a^2+bc=1$
The last one must be the case where you're stuck, and we can try:
$a^2+bc=1\Longleftrightarrow bc=(1-a)(1+a)\Longrightarrow $
$(i)\;\;\;a=0\;\Longrightarrow bc=1\Longrightarrow b=c=1\,\,or\,\,b=c= -1$ \ $(ii)\;\;\;a\neq 0\Longrightarrow \text{ a pair of possibilities for each solution } b\,,\,c\,\,\text{with different signs}$
For example
$A=\begin{pmatrix}\;\;3&\;\;4\\-2&-3\end{pmatrix}$
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0Just to state the hopefully obvious: In case (ii), you can have more than two $b,c$ pairs, depending on how $a^2-1$ factors. E.g. when $a = 5$ you can have -4 and 6 but also 3 and -12; so the solutions for $b,c$ are in this case not necessarily of the form $1+a$, $1-a$. Also, you or the OP may need to treat the case (3) and $a=1$ separately! – 2012-09-04