Are there any parametric solution for this Pythagoras type quadratic Diophantine equation?

Question

I wont to find a parametric solution for the Diophantine equations
$x^2+y^2=m(m+1)$ and
$x^2-y^2=m(m+1).$
I can simplify them up to
$(2x)^2±(2y)^2+1=(2m+1)^2,$ but after that I'm stuck. How can I obtain a parametric solution for this type of equations?

Answer 1

Perhaps you should write a more General solution of the equation.

$$x^2+y^2=z(z+a)$$

If you ask any integers: $k,s$ - and use the solutions of the equation Pell.

$$p^2-2k(k-s)t^2=a$$

The solution, you can write:

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

$$y=p^2-(3k-s)pt+k(2k-s)t^2$$

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

$$.......$$

$$x=(kp+s(s-k)t)t$$

$$y=((k-s)p+kst)t$$

$$z=(2k^2-2ks+s^2)t^2$$

You can write a simple solution without the Pell equation.

If: $a=2t+1$

$$x=b$$

$$y=b^2+t(t+1)$$

$$z=b^2+t^2$$

If: $a=2t$

$$x=b$$

$$y=\frac{b^2+t^2-1}{2}$$

$$z=\frac{b^2+t^2+1}{2}-t$$