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I need to find how many solutions there are, when $A=A^{-1}$ and $A$ is $2 \times 2$ in $\mathbb{Z}_{26}$. I know that $\det(A) \equiv \pm 1 \pmod {26}$. Hence I have $$ \begin{bmatrix} a & b\\ c & d\\ \end{bmatrix} = \det(A)^{-1} \begin{bmatrix} d & -b\\ -c & a\\ \end{bmatrix} $$ $$\det(A)=ad-bc$$ I use Chinese Reminder Theory to find how many solutions there are in $\mathbb{Z}_{2}$ and $\mathbb{Z}_{13}$. Then I can find how many solutions there are overall in $\mathbb{Z}_{26}$. But I run into a problem when I consider this in $\mathbb{Z}_{13}$, when $\det(A)\equiv-1\pmod{13}$. From matrix equivalence I have that $$\begin{cases} a=-d\\ b=b \\ c=c \\ d=-a\\ ad-bc \equiv -1 \pmod{13} \end{cases}$$. How can I find how many solutions there are for equation $ad-bc \equiv -1 \pmod{13}$ or $-a^2-bc \equiv -1 \pmod{13}$?

mle
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Herman
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  • Solutions... to what ?? – DonAntonio Mar 09 '14 at 13:59
  • Sorry for being amiguous. I need to find how many matrices there are that holds $A=A^{-1}$. Just a count no need for particular matrices. – Herman Mar 09 '14 at 14:11
  • but then why $;\det A=\pm 1;$ ? The determinant must be a unit modulo $;26;$ for the matrix to be invertible...or is it another condition? – DonAntonio Mar 09 '14 at 14:37
  • It is given that determinant can be 1 or -1 in modulo 26 and it follows the same for modulo 13. I just need a way to find how many unique solutions there are for $-a^2-bc \equiv -1 (mod13)$. – Herman Mar 09 '14 at 15:28
  • I don't get it: the determinant.."can" be $;\pm1;$ ir else it is given that $;A=A^{-1};$ and $;\det A=\pm1;$ ? – DonAntonio Mar 09 '14 at 15:40
  • I had to prove it that determinant is always 1 or -1 when $A=A^{-1}$, because $A*A^{-1}=I_n$ and $det {I_n}= \pm 1$, hence $det(A)= \pm 1 (mod26)$. And Im allowed to use it for finding a number of possible unique matrices that holds equality $A=A^{-1}$. I consider it in modulo 2 and modulo 13 (CRT) to find how many solutions there are in modulo 26. Also I consider these two cases with both $det(A)=1$ and $det(A)=-1$. And thats when I run into problem, because I get $-a^2-bc \equiv -1 (mod13)$ and don't know how to find the count of possible $a,b,c,d$ sets that would hold for $A=A^{-1}$. – Herman Mar 09 '14 at 15:52

1 Answers1

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We seek the cardinality of the matrices $A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}\in M_2(\mathbb{Z}_{13})$ s.t. $a^2+bc=1$.

Case 1: $c\not=0$, then $b=\dfrac{1-a^2}{c}$, that is $13.12=156$ solutions.

Case 2: if $c=0$, then $a=\pm 1$, that is $13.2=26$ solutions.

The required cardinality is $156+26=182$ non-trivial solutions.

We must add the $2$ obvious solutions $\pm I_2$ and there are $184$ matrices $A\in M_2(\mathbb{Z}_{13})$ s.t. $A^2=I_2$.