<s>[INST] 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?
 [/INST] Since OP said "I want integer solutions for this," meaning $b, c,$ and $d$ should be integers. In that case, $c$ can only be $1$ or $-1$ since the second eqiation implies:
$$c(b−d) = −1  \tag{i}$$
Also, in either case of $c = \pm 1$, the first eqiation implies:
$$b + d = (\pm 1)^2 - 6 = -5 \tag{ii}$$
From the equation $(i)$, you find that if $c=+1$, $b = d−1$ and when $c=-1$, $b = d+1$. Thus, for in general, when $c=\pm 1$:
$$b = d \mp  1\tag{iii}$$
If you apply these findings on the third eqiation, you can obtain numerical values for  $b,d$. For example: When $c=+1$,  $b = d−1$ and if you substitute those values on the third eqiation:
$$bd = d(d-1) = d^2 - d = 6 \ \Rightarrow \  d^2 - d - 6= 0$$
Factors for that equation is $(d - 3)(d + 2) = 0$.
As a result, $d=+3 $ or $d=-2 $. If $d=+3 $, then $b=+3 -1 = +2$. If $d=-2 $, then $b=-2 -1 = -3$.
Now check these two sets $(b,d = 2,3$ and $b,d = -3,-2)$ with second equation:
$$b-d = \frac{-1}{c}  \ \Rightarrow \  2-3 = \frac{-1}{+1} = -1$$
$$b-d = \frac{-1}{c}  \ \Rightarrow \  -3-(-2) = \frac{-1}{+1} = -1$$
Thus both sets satisfied $c=1$ conditions. Therefore, half of solutions for $(b,d,c)$ set is $(2,3,1)$ and $(-3,-2,1)$.
You can solve for other two sets using $c= -1$ easily.

Following is my solution for other two sets using $c= -1$:
When $c=-1$,  $b = d+1$ and when we substitute those values on the third eqiation:
$$bd = d(d+1) = d^2 + d = 6 \ \Rightarrow \  d^2 + d - 6= 0$$
Factors for above equation is $(d + 3)(d - 2) = 0$.
As a result, $d=-3 $ or $d=+2 $. If $d=-3 $, then $b=-3 +1 = -2$. If $d=+2 $, then $b=+2 +1 = +3$.
Now check these two sets $(b,d = -2,-3$ and $b,d = +3,+2)$ with the second equation:
$$b-d = \frac{-1}{c}  \ \Rightarrow \  -2-(-3) = \frac{-1}{-1} = +1$$
$$b-d = \frac{-1}{c}  \ \Rightarrow \  +3-(+2) = \frac{-1}{-1} = +1$$
Thus both sets satisfied $c= -1$ conditions. Therefore, other half of solutions for $(b,d,c)$ set is $(-2,-3,-1)$ and $(+3,+2,-1)$.
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