How to find $x$ such that $a^2-b^2-(c+d)x=r^2$; $c^2-d^2+(a+b)x=s^2$

Question

Let $a,b,c,d,r,s,x$ be integers. Find all integers $x$ such that the following quantities are integral squares: $$a^2-b^2-(c+d)x=r^2$$ $$c^2-d^2+(a+b)x=s^2$$ First, I noticed that if $$(a+b)(c+d)=0$$ then it is easy to solve. I supposed $(a+b)(c+d)\neq 0$. I tried to compare the 2 values of $x$, I did not get too far. Please help. Thanks.

Answer 1

It should be obvious that if $a^2- b^2- (c+ d)x= r$ then $x= \frac{a^2- b^2- r}{c+ d}$. And if $c^2- d^2- (a- b)x= s$ then $x= \frac{c^2- d^2- s}{a- b}$.

In order that x satisfy both of those equations, it must be true that $\frac{a^2- b^2- r}{c+ d}= \frac{c^2- d^2- s}{a- b}$.

Answer 2

Eliminating $x$, we get an equation

$$ (a+b) r^2 + (c+d) s^2 = (a-b)(a+b)^2 +(c-d)(c+d)^2 $$

If $a+b $ and $c+d $ have the same sign, there are finitely many integer solutions; if the signs are different, there may be infinitely many, but they can be parametrized (as a Pell-type equation).

You then look for integer solutions such that $r^2 - a^2 + b^2$ is divisible by $c+d$.

EDIT: For example, if $a=5, b=-3, c=5, d = -1$, the equation in $r$ and $s$ is $$ 2 r^2 + 4 s^2 = 128$$ whose integer solutions are $r = \pm 8$, $s=0$. This does have $r^2 - a^2 + b^2 = 48$ divisible by $c+d=4$, and we get $x = -12$.

Answer 3

For the system of equations.

$$\left\{\begin{aligned}&A^2-B^2-(C+D)X=R^2\\&C^2-D^2+(A+B)X=S^2\end{aligned}\right.$$

Make such a change.

$$a=x(t+p+z-x)+z(t+p-k)+k(k-t-p)+p^2-t^2$$

$$d=x(t+z-p-x)+z(t-p-k)+k(k-t+p)-(p-t)^2$$

$$b=x(p+z-t-x)+z(p-t-k)+k(k+t-p)+(p-t)^2$$

$$s=x(t+p+z-x)+z(t+p-3k)+k(3(k-p)+t-2x)+p^2-t^2$$

$$r=x(3(t-x)+2k-z-p)+z(t+p-k)+k(k-t-p)+p^2-t^2$$

$$c=x(t+p-z-x)+z(3t+k-p-2z)+k(k-t-p)+p^2-t^2$$

The solution can be written as.

$$A=r(r-c-d)+c(2a+s-c-d)+d(a+s)-a(s+r)+b^2$$

$$B=b(a+c-s-r)$$

$$X=(a-c+s-r)(a+c-s-r)$$

$$C=r(r-c-d)+c(2a-r-d)+d(a+s)+b^2-a^2$$

$$D=d(a+c-s-r)$$

$$R=r(a-s-d)+c(a+s-c-d)+d(a+s)+b^2-a^2$$

$$S=s(c-b-r)+r(a+b)-a(a+b-c)+d^2+bc-c^2$$