Diophantine Equation $x^2+y^2=2\cdot z^2$ in terms of the correct solution of the Pythagorean Equation

 In this post we will solve in integers the equation

$$x^2+y^2=2\cdot z^2 \tag{E}$$

Noting that  $(x,y,z)=(\pm a,\pm b,\pm c)$ are solutions, once one of them is, we will limit ourselves to solutions in positive integers. 

We will see that the solution of equation (E) depends on the general solution of the Pythagorean equation $x^2+y^2=z^2$. But I have explained this in more detail here.

          If the integers $(x,y,z)$ verify the equation (E) then, since  $2\cdot z^2$ is an even number, $x$ and  $y$  must necessarily have the same parity. So $\frac{x+y}{2}$ and  $\frac{x-y}{2}$ are integers. It can be seen that (E) is equivalent to

$$\left ( \frac{x+y}{2}\right )^2+\left ( \frac{x-y}{2} \right )^2=z^2 \tag{P}$$

which shows that  $\left (\frac{x+y}{2},\frac{x-y}{2},z \right )$  are solutions of the Pythagorean equation.

     Returning to the previously mentioned Post, the positive integers solutions (not only the primitive ones) of the equation  (P) are given by the families with parameters $s,\; t$ and $d$ ($d$ being a multiplicity factor):

$\begin {cases}\frac{x+y}{2}=2dst\\\frac{x-y}{2}=d(s^2-t^2) \tag{1}\\z=d(s^2+t^2) \end{cases}$

and

$\begin{cases}\frac{x+y}{2}=d(s^2-t^2)\\\frac{x-y}{2}=2dst \tag{2}\\z=d(s^2+t^2)\end{cases}$

where $d\in \mathbb{Z}$ and

$s>t>0$ are two coprime integers of opposite parity.                        (st)

          Then the general solution in positive integers of equation (E) will be, solving for (1) and (2):

$x=d\cdot (s^2+2st-t^2),\;\;y=d\cdot (-s^2+2st+t^2),\;\;z=d\cdot (s^2+t^2) \tag{3.1}$

and

$x=d\cdot (s^2+2st-t^2),\;\;y=d\cdot (s^2-2st-t^2),\;\;z=d\cdot (s^2+t^2) \tag{3.2}$

We notice that the value of  $x$  in the two formulas is the same. And the value of  $y$  is given by two formulas differing only in sign: for $s>\sqrt{2}\cdot t$  we have $y_1<0$ and for $s<\sqrt{2}\cdot t$  we have  $y_2<0$. Since we are only interested in positive values, we can express the final result in the form

          The integer and positive solutions of the equation (E) are

$x=d\cdot (r^2+2st-t^2),\;\;y=d\cdot |s^2-2st-t^2|,\;\;z=d\cdot (s^2+t^2) \tag{DST}$

where $d\in \mathbb{N}$  and  $s,\;t$  check the condition (st) above.

          Remark CiP  In the table below we show some solutions for $d=1$ (i.e. primitive solutions, gcd(x,y,z)=1)

ATTENTION! The values ​​$s=9,\; t=7$  do not satisfy the condition (st). The solution $(x,y,z)=(158,94,130)=2\cdot (79,47,65)$ is the multiple of another solution...