Solving the Diophantine system of equations $w + x + y - z = a, wx + yz = b$

Question

If we have a Diophantine system of equations of the form

$$ \begin{align*} x + y & = a, \\ xy & = b \end{align*} \tag{1} $$

with $x,y$ unknown and $a,b$ known, we could solve it using the identity

$$ \begin{align} (x-y)^2 & = (x+y)^2 -4(xy) \\ & = a^2 - 4b \end{align} \tag{2} $$

The Diophantine system of equations has a solution iff $a^2 - 4b = k^2$ is a perfect square, for some integer $k$ where $k+a$ is even. We can then solve $x-y = k, x+y=a$ to solve for $x, y$.

While working on a problem, I have obtained a system of equations of the following form:

$$ \begin{align} w + x + y - z & = a \\ wx + yz & = b \end{align} \tag{3} $$

This looks tantalizingly close to the form of Eqn. (1) with a twist and I am not able to proced further.

Is there a way to find parametric solutions for $w, x, y, z$ in Eqn. (3)?

Answer 1

Comment: Does this idea help you:

For example we want to solve following system of equations:

$\begin{cases} x+w+y-z=2\\xw+yz=14\end{cases}$

Suppose:

$w=x+d_1, y=x+d_2, z=x+d_3$

we have:

$x+w+y-z=x+(x+d_1)+(x+d_2)-(x+d_3)=2x+(d_1+d_2-d_3)=2$

$x(x+d_1)+ (x+d_2)(x+d_3)=2x^2+(d_1+d_2+d_3)x+d_2\cdot d_3=14$

So problem reduces to:

$\begin{cases}2x+(d_1+d_2-d_3)=2\\2x^2+(d_1+d_2+d_3)x+d_2\cdot d_3=14\end{cases}$

take for example $x=1$ , we have:

$\begin{cases}(d_1+d_2-d_3)=0\\d_1+d_2+d_3+d_2\cdot d_3=12\end{cases}$

So we get:

$2d_3+d_2d_3=d_3(2+d_2)=12=3\times 4=2\times 6$

take for instance $d_3=3$, this gives $d_2=2$ and $d_1=1$; so solutions are:

$(x, w, y, z)=(1, 2, 3, 4)$

Answer 2

$wx+yz=b$

$w+x+y-z=a$

Above simultaneous equation has more than one

parametrization:

For '$w$'=$12$, we have:

$(a,b)=[(12m),(m^3)]$

$(x,y,z)=[(m^2+11m-132),(132-m^2),(12-m)]$

And for $m=8$ we get numerical solution:

$(w,x,y,z,a,b)=(12,20,68,4,96,512)$