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I am trying to find solutions for this diophantine equation

$$x^2+y^2+x^2y^2=4z^2$$

I am looking for advice on a procedure to find all positive integer solutions for this equations.

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    $\begingroup$ Add $1$ to both sides and you get $(x^2+1)(y^2+1)=4z^2+1$, which is only possible if $x$ and $y$ are even, which reduces to another problem posted earlier today: $(4x^2+1)(4y^2+1)=4z^2+1$. math.stackexchange.com/q/162862/7933 $\endgroup$ – Thomas Andrews Jun 25 '12 at 17:54
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by considering x,y are both even and let $x^2=s ,y^2=r$

then $sr+s+r=(2z)^2$ this equation is equivalent to:

$(2s+r+1)^2=(r-1)^2+(2s)^2+(4z)^2$ you can check that and the positive solutions for the last equation are given by the dimensions and the length of the diagonal of a rectangular box which is a related problem to Pythagorean Triple.

so $s=a , b=a+1 , r=(4z^2-a)/b$

& b is a divisor of $a^2+4z^2 ,b<√(a^2+4z^2 ) , 1<2z^2$

one can solve this equation over the positive integers if he just know the value of $a$

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  • $\begingroup$ This is not a solution. $\endgroup$ – individ Aug 14 '16 at 14:02
  • $\begingroup$ so what is it ? then Pythagorean solution is not a solution? $\endgroup$ – Mahmoud. A .Solomon Aug 14 '16 at 14:07

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