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is it 2xy-z^2=0??

oh wait , forgot to type it out, no, 2xy-z^2=1

one such point is (1,1,1)

only this is the solution..no other points are there.

got it??

i get that z can be either -1 ,1 but how did you get x^2+y^2+z^2=3?

and also from second approach z^2<=1
z^2=1
z=1,-1
so 2xy=2
xy=1
and both x,y>0
so x=y=1

2xy=z^2+1
(x+y)^2=4
x^2+y^2+2xy=4
x^2+y^2+z^2=3