Nonlinear diophantine equations $x^2+2y=z^2$ and $y^2+2x=w^2$ I am asked to find two sets of positive numbers $x$ and $y$, such that both $x^2+2y$ and $y^2+2x$ are perfect squares. 
I found a general solution to either single equation, but it seems impossible to satisfy them both. Can it be done, and why or why not?

UPDATE: What if $x$ and $y$ are positive rational numbers?
 A: $x^2 + 2y = z^2 \to x^2 < z^2 \to x < z \to x+1 \leq z \to (x+1)^2 \leq z^2$
$y^2 + 2x = w^2 \to y^2 < w^2 \to y < w \to y+1 \leq w \to (y+1)^2 \leq w^2$.
Thus: $(x+1)^2 + (y+1)^2 \leq z^2+w^2$ (*)
But:  $x^2+2y+y^2+2x = z^2 + w^2 \to (x+1)^2+(y+1)^2 = z^2+w^2+2 > z^2 + w^2$ (**).
(*) and (**) contradict each other. So there is no natural solutions.
A: It is quite famous and old task. She is still in the book of Diophantus described.
On this forum I solution stirred, but the topic was deleted.
This system of equations solved before Diophantus.  $$\left\{\begin{aligned}&x^2+ty=z^2\\&y^2+tx=q^2\end{aligned}\right.$$
Although elementary obtained such solutions:  $$x=b-a$$  $$y=b+a$$  $$t=8b$$  $$z=3b+a$$  $$q=3b-a$$  
more:  $$x=a-3b$$  $$y=a+b$$  $$t=8(a-b)$$  $$z=3a-b$$  $$q=3a-5b$$
more:  $$x=p^2-10ps-11s^2$$  $$y=2p^2+4ps+38s^2$$  $$t=12(2p+5s)s$$  $$z=p^2+14ps+49s^2$$  $$q=2p^2+10ps-28s^2$$  
more:  $$x=p^2+10ps+21s^2$$  $$y=2p^2+12ps+22s^2$$  $$t=-4(2p+5s)s$$  $$z=p^2+2ps+s^2$$  $$q=2p^2+10ps+8s^2$$  
But these solutions are not of interest. The fact that a decision that leads Diophantus described sleduyushy formula.  $$x=2psb^2-a^2p^2$$  $$y=2abp^2-b^2s^2$$  $$t=as(4bp-as)$$  $$z=2psb^2+a^2p^2-abs^2$$  $$q=2abp^2+b^2s^2-psa^2$$  
I do not think that Diophantus accidentally brought this decision. But then he had to know this formula. Although the cause could not be more simple. And specifically chose not even solution.
If there is some factor before $t - $ the formula enough to divide by this number. This entry solution is better because it allows you to have a more General formula.
