A system of Diophantine equations

Question

Is there any positive solution for the Diophantine system $$\left\{\begin{array}{l} x^2-4y=z^2 \\ y^2-4x=w^2 \end{array} \right. ,$$ except that $(x,y)=(4,4), (6,5)$ (up to symmetry)?

Answer 1

$x^2-4y=z^2$

$y^2-4x=w^2$

$x^2-y^2-4(y-x)=z^2-w^2$

$(x-y)(x+y+4)=(z-w)(z+w)$

Since factors in LHS and also in RHS of this relation are co-prime then we can write:

$x-y=z-w$

$x+y+4=z+w$

$⇒ 2x+4=2z$ $⇒z=x+2$

$2w=2y+4$ $⇒w=y+2$

Plugging z or w in 1st or 2nd equation we get:

$x+y=-1$

Therefore:

$z+w=x+y+4=-1+4=3$

Now we make following table for one set of possible solutions:

$z= 0, 1, 2, 3$

$w= 3, 2, 1, 0$

$x= -2, -1, 0, 1$

$y= 1, 0,-1, -2$

Now if $x-y=1$ then we get:

$x^2-y^2 +4=z^2-w^2$

or: $x^2+w^2 +4 = z^2+y^2$

Which gives: $x=6, w=1, z=4, y=5$

If $y-x=1$ then:

$y^2-x^2 +4=w^2-z^2$

Which gives: $y=6, z=1, w=4, x=5$

If $z=w=0$ then $x=y=4$

These are only possible sets of positive solutions.