A in M2(prime), A=B^2 and det(B)=p^2

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A in M2(prime), A=B^2 and det(B)=p^2

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#1 h May 31, 2021, 5:52 AM • 2 Y

Y by centslordm, Mango247

Determine all $2\times2$ integer matrices $A$ having the following properties:

$1.$ the entries of $A$ are (positive) prime numbers,
$2.$ there exists a $2\times2$ integer matrix $B$ such that $A=B^2$ and the determinant of $B$ is the square of a prime number.

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KAME06

#2 h Apr 22, 2025, 9:59 PM

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Suppose that a matrix $A$ has the properties. Consider $B= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . Then:
$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a^2+bc & ab+bd \\ ac+dc & bc+d^2 \end{pmatrix}=\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix}$ .
We know that $b(a+d), c(a+d)$ are primes, so we have two cases:
Case 1: $a+d= \pm 1 \Rightarrow d= \pm1-a \Rightarrow A=\begin{pmatrix} a^2+bc & \pm b \\ \pm c & (1-a)^2+bc \end{pmatrix}$ .
All entries are prime, so $a^2+bc \ge a^2+(2)(2) \ge 4$ . Analogous, $(1-a)^2+bc \ge 4$ , so both might be odd, but $a^2$ and $(1-a)^2$ don't have the same parity, so $(1-a)^2+bc$ and $a^2+bc$ can't be odd. Contradiction.
Case 2: $b=c= \pm 1$
We know that $det(B)=ad-bc=ad-1=p^2$ where $p$ is a prime, then $ad=p^2+1>0$ so $a, d$ aren't $0$ and have the same sign, so WLOG $b=c=1$ . Then
$A=\begin{pmatrix} a^2+1 & a+d \\ a+d & d^2+1 \end{pmatrix}$
If $3 \mid p \Rightarrow ad \not \equiv 1 \pmod{3}$
--WLOG, if $3 \mid a$ , then $p^2 =ad-1 \equiv -1 \pmod{3}$ , but $-1$ isn't quadratic residue mod 3. Contradiction.
--WLOG, if $a \equiv 1 \pmod{3}$ and $d \equiv 2 \pmod{3}$ , then $3 \mid a+d$ but $a+d$ is prime, so $a+d=3 \Rightarrow (a, d)=(1, 2)$ but $det(B)=(1)(2)-1=1 \neq p^2$ . Contradiction.
If $3 \mid p$ , then $p=3 \Rightarrow ad=3^2+1=10 \Rightarrow (a, d)=(10, 1); (1, 10); (2, 5); (5,2)$ , but note that $5^2+1=26$ and it isn't prime. So we have $(a, b, c, d)=(10, 1, 1, 1); (1, 1, 1, 10)$ , then $A \in \left \{ \begin{pmatrix} 101 & 11 \\ 11 & 2 \end{pmatrix},\begin{pmatrix} 2 & 11 \\ 11 & 101 \end{pmatrix} \right \}$
Checking that $101, 11, 2$ are primes, $\begin{pmatrix} 101 & 11 \\ 11 & 2 \end{pmatrix}=\begin{pmatrix} 10 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 10 & 1 \\ 1 & 1 \end{pmatrix}$ , $\begin{pmatrix} 2 & 11 \\ 11 & 101 \end{pmatrix}=\begin{pmatrix} 1 & 1 \\ 1 & 10 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 10 \end{pmatrix}$ , and $(1)(10)-(1)(1)=9=3^2$ , we conclude that the answers are $\begin{pmatrix} 101 & 11 \\ 11 & 2 \end{pmatrix},\begin{pmatrix} 2 & 11 \\ 11 & 101 \end{pmatrix}$ .

This post has been edited 1 time. Last edited by KAME06, Apr 22, 2025, 10:00 PM

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