Making An Equation Complex Doesn't Make It Harder

Since $(x,y)\neq(0,0)$ , we can set $x=\rho\cos\theta,y=\rho\sin(\theta)$ with $\rho>0,\theta\in[0,2\pi)$ and solve: $1 = \rho^3 (\sin\theta-\cos\theta) = \rho^4\left(\frac{3}{\sqrt[3]{2}}+2\rho^2\sin\theta\cos\theta \right).$ From the first equation we get $\sin\theta-\cos\theta = \sqrt{2}\sin\left(\theta-\pi/4\right) = \frac{1}{\rho^3}$ , and by squaring we have $2\sin\theta\cos\theta = 1-\frac{1}{\rho^6}$ . Substituting this identity into the second equation we have: $\rho^6+\frac{3}{\sqrt[3]{2}}\rho^4-2=0,\qquad \sin\left(\theta-\pi/4\right)=\frac{1}{\sqrt{2}\cdot\rho^3}.$ Now setting $u = \rho\cdot\sqrt[6]{2}$ we have: $u^6+3u^4-4=(u^2-1)(u^2+2)^2=0,\qquad \sin\left(\theta-\pi/4\right)=\frac{1}{u^3}.$ Now the only positive solution of the left-hand equation is clearly just $u=1$ , so we have $\theta=3\pi/4$ and just one solution $(x,y)$ .

Denote: $t=y-x \Rightarrow x^{2}+(t+x)^{2}=\frac{1}{t} \Rightarrow \Delta=\frac{2}{t}-t^2 \geq 0 :(1)$ $y-x=(2xy+\frac{3}{\sqrt[3]{2}})(x^2+y^2) \Rightarrow 2x(x+t)+\frac{3}{\sqrt[3]{2}}-t^{2}=0$ $\Rightarrow \Delta = 3t^{2}- \frac{6}{\sqrt[3]{2}} \geq 0: (2)$ $(1), (2) \Rightarrow t=\sqrt[3]{2}$ $\Rightarrow$ there is one $\boxed{1}$ solution

Great solutions presented above. Let me explain the title of the problem, and also how the strange values were chosen.

Clearly $x = y = 0$ is not a solution.
If $x + y = 0$ , then we can easily verify that the only solution is $x = - \frac{ 1 } { \sqrt[3]{4}}$ , $y = \frac{ 1 } { \sqrt[3]{4} }$ .

Otherwise, we have $x+y \neq 0$ . Set $z = x+ iy$ . We have
$x^2 - y^2 = (x+y) ( x-y) = \frac{ x+y} {x^2 + y^2 }$ and $2xyi + \frac{3} { \sqrt[3]{2} } i = \frac{ y-x} { x^2 + y^2}$

Adding these two equations up, we get $x^2 + 2xyi - y^2 + \frac{ (1+i) ( x - iy) } { x^2 + y^2 } + \frac{ 3} { \sqrt[3]{2} } i = 0$ , or that $z^2 + \frac{ (1+i) } { z} + \frac{3} { \sqrt[3]{2} } i = 0$ .

Multiplying throughout by $z \neq 0$ , we get the cubic equation

$z^3 + \frac{ 3} { \sqrt[3]{2} } iz + (1+i) = 0$

This can be factorized as $\left( z + (-1+i) \sqrt[3]{2} \right) \left(z + (1-i) \frac{1}{\sqrt[3]{4} } \right)^2 = 0$ . Observe that both of these roots satisfy $x +y = 0$ , hence they do not satisfy the assumption that $x + y \neq 0$ . Thus, there are no solutions in this case.