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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.

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up vote 5 down vote accepted

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


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Thanks, that's what I meant but expressed poorly. I'll edit. – DonAntonio Sep 4 '12 at 13:19
Looks good. $~$ – Jack Schmidt Sep 4 '12 at 13:49
Just 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! – yatima2975 Sep 4 '12 at 15:16

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