Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$ Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?

By substracting the two equations you get
$$ x^2(1-x^2) + y^2(y^2-1) = 0$$
$$ -(x^2-\frac12)^2 + (y^2-\frac12)^2 = 0 $$
so
$$ x^2 - \frac12 = \pm (y^2-\frac12) $$
that is
$$ x^2 = y^2 $$
or $$ x^2 = 1- y^2 $$
Analysing theses two cases it's easy to solve the equations.
