TITLE: Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$
QUESTION [14 upvotes]: 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?

REPLY [18 votes]: There is actually a much, much neater method. Following your solutions, we subtract the equations (or substitute $20$) to get,
$$(u-v)+(v^2-u^2)=0$$
Using the familiar identity $a^2-b^2$ $=(a-b)(a+b)$,
$$-(v-u)+(v-u)(v+u)=0$$
which, on factoring, implies
$$(v-u)(-1+v+u)=0$$
which implies
$$v=u\text{ or }v=1-u$$
All that remains is mere substitutions into our original equations.
Hope this helps. Have a wonderful day. Ask anything if not clear :)