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