How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations and substituting into the first. None of these methods worked. How do I do this?

From the first two equations we get $b,d$ in terms of $c$:
$$b=\frac12\left(c^2-6-\frac{1}{c}\right)$$
$$d=\frac12\left(c^2-6+\frac{1}{c}\right)$$
Now substitute into the third:
$$\frac14\left((c^2-6)^2-\frac{1}{c^2}\right)=6$$
Let $a=c^2$:
$$(a-6)^2-\frac{1}{a}=24$$
$$a^3-12a^2+12a-1=(a-1)(a^2-11a+1)=0$$
Since $c$ is integral, so is $a$ and hence $a=1$ and $c=\pm1$; setting $c=1$ leads to $b=-3$ and $d=-2$ while $c=-1$ gives $b=-2$ and $d=-3$. Thus the integer solutions are
$$(b,c,d)=(-3,1,-2)\lor(b,c,d)=(-2,-1,-3)$$
