Squaring Around

multiply them and you get $xy=1$

in any eqn you multiply x or y , you get respectively $xy=y^3$ and $xy=x^3$ .

Now except 0 and 1 you think (Obvoiusly those two are correct too.) . $x^3=1$ , it has 3 values which satisfy the eqn (so for $y^3=1$ ). To find them , Let's factorise ,

$x^3-1=(x-1)(x^2+x+1)$

$x-1=0$ leads $x=1$ (which we have before) , and $x^2+x+1 =0$ renders us 2 complex roots traditionally they are expressed as $\omega , \omega^2$ .

$\displaystyle \omega = \frac{-1+\sqrt{3}i}{2}$

$\displaystyle \omega^2 = \frac{-1\sqrt{3}i}{2}$

There we get 4 solutions , $(x,y)=(0.0).(1,1),(\omega,\omega^2),(\omega^2,\omega)$

All can be written as $a+bi$ , so 4 is the answer . (Though I missed taking 0,1 for it . )