A system of Diophantine equations 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)?

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