<s>[INST] How many real solutions $(x,y)$ for $x + y^2 = y^3 $ and $y + x^2 = x^3$ I think the answer is that there are $3$ real solutions.


*

*When $x = y$

*When $x = -y$
If $x\neq y$ is there a real solution? If so, how do we find the real solutions?
 [/INST] Subtracting one equation from the other, we obtain $$(x-y)+\left(y^2-x^2\right)=\left(y^3-x^3\right)\Rightarrow$$ $$(x-y)\left(1+(x+y)+\left(x^2+xy+y^2\right)\right)=0\Rightarrow$$ $$(x-y)\left(\frac14 (1 + 2 x + y)^2+\frac1{12}(3y+1)^2+\frac23\right).$$ The second factor is strictly positive, so the only possibility is that $x=y$. In this case, our initial condition becomes $$x+x^2=x^3,$$ which gives the $\boxed{3}$ solutions $$(0,0),\left(\frac{1-\sqrt5}2, \frac{1-\sqrt5}2\right),\left(\frac{1+\sqrt5}2, \frac{1+\sqrt5}2\right).$$
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