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?
 A: If we call $ \ b \ $ and $ \ d \ $ the roots of a quadratic equation, the Viete relations tell us that the equation is $ \ x^2 + (6 - c^2)x + 6 \ = 0 \ \ \ $ (from the sum and product of the roots).  The difference of the zeroes of a monic quadratic polynomial is equal to the square-root of its discriminant $ \ ( \ b - d \ = \ \sqrt{D} \ ) \ \ , \  $ so we have
$$ (b \ - \ d)^2 \ \ = \ \ (6 - c^2)^2 \ - \ 4·1·6 \ \ = \ \ c^4 \ - \ 12c^2 \ + \ 12 \ \ = \ \ \frac{1}{c^2} \ \ .  $$
This leads to the apparently unpleasant sixth-degree equation $ \ c^6  -  12c^4  +  12c^2  - 1 \  = \ 0 \ \ . \ $  But the polynomial is an even function and (anti-)palindromic, so both $ \ c \ = \ 1 \ $ and $ \ c \ = \ -1 \ $ are among its roots, leading to the factorization $ \ (c^2 - 1)·(c^4 \ - \ 11c^2 + 1) \ \ , $
using polynomial or synthetic division.
Since the second factor is a biquadratic polynomial, it is not hard to calculate using the quadratic formula that none of its zeroes are integers (the Rational Zeroes Theorem also tells us this).  We thus have only one possibility for obtaining integer roots of our original equation
$$ \mathbf{c \ = \ \pm 1 \ \ : } \quad \quad x^2 + 5x + 6 \ = \ (x + 3)·(x + 2) \ = \ 0 \ \ \Rightarrow \ \ b \ , \ d \ \ = \ \ -3 \ , \ -2 \ \ . $$
$$ \ \ $$
ADDENDUM --
We can also interpret this system geometrically.  The equation $ \ xy \ = \ 6 \ $ describes a (rectangular) hyperbola with its major axis on the line $ \ y \ = \ x \ \ . \ $  The line $ \ x - y \ = \ -\frac{1}{c} \ $ runs parallel to this major axis, but we are not permitted $ \ c \ = \ 0 \ \ . \ $ There will be just two intersections of such a line with the hyperbola, one in the first and one in the third quadrant; thus $ \ b \ $ and $ \ d \ $ must have the same sign.  Since we wish both of these numbers to be integers, the only possibilities are those laid-out in marty cohen's answer:  the "diagonal" symmetry of the hyperbola allow us to write these as
$$ (b \ , \ d) \ \ = \ \ (\pm 1 \ , \pm 6) \ \ , \ \ (\pm 2 \ , \pm 3) \ \ \ \text{[either permutation].} $$
These correspond to the lines $ \ x - y \ = \ \pm 5 \ \ $ or $ \ x - y \ = \ \pm 1 \ \ , \  $ respectively.
But these points must also lie on a line $ \ x + y \ = \ c^2 - 6 \ \ge \ -6  \ \ .  \ $  This eliminates the pair $ \ (-1 \ , \ -6) \ \ . \ $ As for the others, $ \ (1 \ , \ 6) \ $ requires $ \ x + y \ = \ 7 \   \Rightarrow \ c \ = \ \pm \sqrt{13}   \ \ ,  \ $ which is inconsistent with the difference equation; similarly, $ \ (2 \ , \ 3) \ \Rightarrow  \ x + y \ = \ 5 \   \Rightarrow \ c \ = \ \pm \sqrt{11}   \ \ .  \ $  The remaining pair $ \ (-2 \ , \ -3) \ $ produces  $ \ x + y \ = \ -5 \   \Rightarrow \ c \ = \ \pm \sqrt1   \ \ ,  \ $ giving us the sole consistent solution to the given system of equations.

$$ \text{rejected solutions} $$

$$ \text{only permissible solutions lie on green line} $$
A: 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
$$(b,c,d)=(-3,1,-2)\lor(b,c,d)=(-2,-1,-3)$$
A: If $b, c, d$ are integers,
then $c$ can only be
$1$ or $-1$.
In either case,
$b+d=-5$.
From the last,
$(b,d)$ is one of
$(1,6), (2,3),
(3,2), (6,1)$,
or these replaced by their negatives.
The only ones that work are
$(-2,-3)$ and $(-3,-2)$.
If $c=1$, then $b-d=-1$ so
$b=-3$, $d=-2$.
If $c=-1$, then $b-d=1$ so
$b=-2$, $d=-3$.
A: 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 $(+3,+2,-1)$.
