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Consider this equation $$z^3\sqrt{x^2+y^2-z^2}=w(x^2+y^2)$$ where $w$, $z$ are positive integers, $x$, $y$ are any integers.

Squaring both sides we get the following Diophantine equation $$z^6(x^2+y^2-z^2)=w^2(x^2+y^2)^2$$ Some solutions in non-negative integers are

(0, 25, 15, 108)
(0, 25, 20, 192)
(2, 2, 2, 2)
(2, 11, 5, 10)
(2, 11, 10, 40)
(2, 14, 10, 50)
(4, 4, 4, 8)
(4, 22, 10, 40)
(4, 22, 20, 160)
(4, 28, 20, 200)
(5, 10, 5, 10)
(5, 10, 10, 40)
(6, 6, 6, 18)
(7, 24, 15, 108)
(7, 24, 20, 192)
(8, 8, 8, 32)
(10, 10, 10, 50)
(10, 20, 10, 40)
(10, 20, 20, 160)

$$\sqrt{x^2+y^2-z^2}=t \implies x^2+y^2=z^2+t^2$$ My approach was to find integers that can be represented as sum of two squares in more than one way. And then check if $w$ is integer. Is it possible to find a parametric solution for this equation? Or maybe an efficient method (other than brute force) to find its solutions? Any help will be appreciated.

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$$z^6(x^2+y^2-z^2) = w^2(x^2+y^2)^2\tag{1}$$

We consider simultaneous equation

$$ \left\{ \begin{array}{c} x^2+y^2 = z^3 \\ x^2+y^2 = z^2+w^2 \\ \end{array} \right. $$

Hence we get $z^2(z-1) = w^2$.

Let $t^2 = z-1$ then $(z,w)=(t^2+1, (1+t^2)t)$.

From first equation, $x^2+y^2 = (t^2+1)^3$.

Hence we get $(x,y)=(3t^2-1, t(-3+t^2))$.

Finally, $(x,y,z,w)=(3t^2-1, t(-3+t^2), t^2+1, (1+t^2)t)$

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$(x,y,z,w)=(u^4 - 6 u^2 v^2 + v^4, 4 u v (u^2 - v^2), 2 u v (u^2 + v^2), 8 u^3 v^3 (u^2 - v^2))$,

where natural $u>v$.

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  • $\begingroup$ Thanks for the reply. A bit of explanation please, if you have time. $\endgroup$ – user125368 Aug 4 '20 at 3:27
  • $\begingroup$ I think you meant $(x,y,w,z)$. Does this give all the solutions? $\endgroup$ – user125368 Aug 4 '20 at 4:18
  • $\begingroup$ Oh yes, sorry, fixed. This is just one of many parametric solutions. $\endgroup$ – Dmitry Ezhov Aug 4 '20 at 6:19

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