Solve $ab + cd = -1; ac + bd = -1; ad + bc = -1$ over the integers

Question

I am trying to solve this problem:

Solve the system of equations \begin{align} \begin{cases} ab + cd = -1 \\ ac + bd = -1 \\ ad + bc = -1 \end{cases} \end{align} for the integers $a$, $b$, $c$ and $d$.

I have found that the first equation gives $d = \dfrac{-1-cd}{a}$, which gives $a\neq0$. Other than that, I don't know where to start.

Tips, help or solution is very appreciated. Thanks!

Answer 1

Add equation 1 and 2, 1 and 3, 2 and 3 you get

$(a+d)(b+c)=-2$

$(a+c)(b+d)=-2$

$(a+b)(c+d)=-2$

All of them are integers, so it can only be $-2\times1$ or $2\times(-1)$.

List out all possibility the solutions can only be any permutation of $(1,1,1,-2)$ or $(-1,-1,-1,2)$

Answer 2

Hint

Subtract equation 2 from equation 1 and equ 3 from equ 1 to get \begin{align*} (a-d)(b-c)&=0\\ (a-c)(b-d)&=0 \end{align*}

From this we can infer that either $a=d$ or $b=c$ and either $a=c$ or $b=d$. Now try these possibilities in the system given. Can you proceed from here?

Answer 3

Hint:

Squaring the equations gives

\begin{align} \begin{cases} a^2b^2 +2abcd+ c^2d^2 = 1 \\ a^2c^2 +2abcd +b^2d^2 = 1 \\ a^2d^2 + 2abcd+b^2c^2 = 1 \end{cases} \end{align}

Subtracting them yields

\begin{align} \begin{cases} (a^2-b^2)(c^2-d^2)=0 \\ (a^2-c^2)(b^2-d^2)=0 \\ (a^2-d^2)(b^2-c^2)=0 \end{cases} \end{align}

Now consider some cases.