Solving $ac-5bd=1$ and $ad+bc=0$

Question

I need to determine the set of values $a,b$ such that the system of equations

$$ac-5bd=1$$

and

$$ad+bc=0$$

has solutions in the integers. I know I can solve the latter for

$$a=-\frac{bc}{d}$$

and get

$$\frac{-bc^2}{d}-5bd=1$$

Clearly exactly one of $b,d$ must be negative, and we can get

$$-bc^2-5bd^2=d$$

which is

$$-b(c^2+5d^2)=d$$

but from here I'm not sure where I could go.

Answer 1

Hint: If $a$ and $b$ are fixed, this is a system of two linear equations in two unknowns, which we can solve to get

$$c=\frac{a}{a^2+5b^2},\ d=-\frac{b}{a^2+5b^2}.$$

When are these both integers?

Answer 2

There are infinitely many solutions. In fact you have $$\begin{pmatrix}a&5b \\ b&a\end{pmatrix}\cdot\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$$ therefore $$x=\frac{a}{a^2-5b^2}\space\space y=\frac{-b}{a^2-5b^2}$$

For example you can do $$a^2-5b^2=-1$$ which is a Pell-Fermat equation whose infinity of integer solutions $(a_n,b_n)$ are given by $$a_n+\sqrt5\space b_n=\left(\frac{1+\sqrt5}{2}\right)^n$$ where $\dfrac{1+\sqrt5}{2}$ is the fundamental unit in the quadratic field $\mathbb Q(\sqrt5)$.