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Let $a,b,c,d$ integers, $c\not=0$ such that $ad-bc=1$ and $c\equiv 0 \pmod p$ for some prime $p>3$.

Show that if $a+d=\pm1$ then $p\equiv 1\pmod 3$

I don't know how to approach this problem because when I take the expression $ad-bc$ modulo $p$ we have that either $d-d^2\equiv 1\pmod p$ or $d+d^2\equiv 1\pmod p$.

Bill Dubuque
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Blacksmith
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3 Answers3

2

Let's consider:

$ad-bc\equiv1 \pmod p\implies ad=1-bc \equiv 1\pmod p$

$(a+d)^2 \equiv a^2+d^2+2ad\equiv a^2+d^2+2\equiv1\pmod p$

$\implies a^2+d^2+1\equiv 0\pmod p$

$\implies (a^2+d^2+1)^2\equiv 0\pmod p$

Since:

$a+d=\pm1 \implies a\cdot d=(k+1)\cdot(-k)=-k^2-k$

Consider the following table $\pmod 3$

$$\begin{array}{ l | c | r } k & a & d & a^2 & d^2 & p \\ -1 & 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 1 & -1 & -1 & 1 & 1 & 0 \\ \end{array}$$

$\implies p \equiv1 \pmod 3 \quad \square$

user
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  • I don't follow the last statement. Why does $ad = 1 \pmod 3$ and $ad = 1 \pmod p$ imply $p = 1 \pmod 3$ – Malcolm Dec 05 '17 at 09:40
  • Also the conclusion that $ad$ is necessarily $1\pmod 3$ is false as the example $a=6$, $b=-1$, $c=31$, $d=-5$ shows – Malcolm Dec 05 '17 at 10:15
  • You are absolutely right, I've just revised the derivation. Do you think it is ok now? Thanks! – user Dec 05 '17 at 13:45
  • How do you arrive at the conclusions about $p\pmod 3$ in your table? Why do you exclude the last row of your table? No, I don't think it's OK now. – Malcolm Dec 05 '17 at 14:11
  • Ok I've understood you objection and you're right, I need to revise my solution. Thanks! – user Dec 05 '17 at 14:17
2

Given $ad-bc=1$ with $c\equiv 0 \pmod p$ for $p$ a prime exceeding $3$. If in addition we have $a+d = \pm 1$ then using $a = \pm 1 - d$ in $ad-bc=1$ $$ 1 \equiv ad - bc \equiv (\pm 1 - d)d - 0 \equiv \pm d -d^2 \quad \text{or}\quad d^2 \equiv \pm d -1 \pmod p $$ $$ d^3 \equiv \pm d^2-d \equiv \pm (\pm d - 1)-d \equiv \mp 1 \implies d^6 \equiv 1 \pmod p. $$ The order of $d \pmod p$ is $1$, $2$, $3$, or $6$. The above also applies to $a$ so the order of $a$ is also restricted to those four possibilities. If we knew that there was an element of order $3$ (or $6$) we would be done because then $3 | (p-1)$.

We need to exclude the cases where the order of $d$ is $1$ or $2$.

If the order of $d$ is $1$ then $d\equiv 1\pmod p$. Using this together with $a+d = \pm 1$ and the observation that $a\not\equiv 0$ gives that $a\equiv -2$. So if the order of $d \pmod p$ is $1$ then the order of $a$ is not $1$ or $2$ so $3 | (p-1)$.

Similarly if the order of $d$ is $2$ then $d\equiv -1$ and $a\equiv 2$ and again $3 | (p-1)$.

An all cases $3|(p-1)$ and $p\equiv 1\pmod 3$.

Malcolm
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1

Here is an advanced solution based on linear algebra, for fun.

Consider the matrix $A=\pmatrix{ a & b \\ c & d}$ over $\mathbb F_p$.

This matrix is upper-triangular, has trace $\pm1$ and determinant $1$. Moreover, $a$ and $d$ are eigenvalues of $A$. Therefore, they are roots of the characteristic polynomial of $A$, which is $X^2\pm X+1$. Thus, its discriminant $-3$ is a square.

Now, $-3$ is a quadratic residue mod $p$ with $p>3$ iff $p \equiv 1 \bmod 6$. See a proof here.

lhf
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