Squares and Integers

Firstly, multiply by $\ x^2 y^2$ to obtain: $\ x^2 +y^2 - z(xy)^2 = 0$ Now there is symmetry in x and y so we would expect the coefficient of $\ x^2 y^2$ to be a square so I shall multiply by z before factorising as follows $\ zx^2 +zy^2 - z^2 (xy)^2 = 0 \Rightarrow\ (zx^2 -1)(zy^2 -1) = 1$ Now $\ 1 = 1 \times\ 1 = (-1) \times (-1)$ so: $\ zx^2 - 1 = -1 \Rightarrow\ xz = 0$ which is impossible as x is on the denominator in the original equation (i.e. $\frac{1}{0}$ is undefined) and z can't equal zero as a reciprocal square is always strictly more than zero. Now for case 2: $\ z x^2 -1 = 1 \Rightarrow\ x^2 = \frac{z}{2} \Rightarrow\ (x,y,z) = ( \pm 1 , \pm 1 , 2 )$ This produces 4 solutions namely $\ (x,y,z) = (1,1,2), (-1,1,2),(1,-1,2),(-1,-1,2)$