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?
 A:  Edited: thanks to several comments below, I find that the original answer made a (serious) mistake in counting the degrees of relevant polynomials and thus a mistaken use of the fundamental theorem of algebra. I apologize for the errors. A not-so-elegant way to fix the argument is added.


Systems of nonlinear equations are in general difficult; solutions are found by numerical methods.
Nevertheless, for this particular problem,
by inspection$\dagger$, there are at least four solutions:
$$
(-2,-2),\quad (-2,2),\quad (2,-2),\quad (2,2)\tag{1}
$$
We claim that the system has at most four solutions.
On the other hand, by symmetries, it suffices to show that $(2,2)$ is the unique solution in the first quadrant: any other solution not in (1) would give you one more solution in the first quadrant.
If you introduce the new variables $u=x^2$ and $v=y^2$:
$$
v=20-u^2,\quad u=20-v^2,\quad u,v>0
$$
The pair $(u,v)$ is the intersection of two parabolas on the $u$-$v$ plane.
If one draws a picture$\dagger\dagger$, one can see that one has only one intersection on the first quadrant, which implies that the first-quadrant solution to the original system is unique.

$\dagger$Notes. The observation $20=4+16$ gives an obvious hint to get a solution.
$\dagger\dagger$ Yes, this is just a hand-waving geometric "proof".

A: 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.
A: 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 :)
A: First, you can guess the answer, which might be $(u, v) = (2, 2)$. Since there might exist other solutions, we need to think more.
The next thing is: I believe that you can draw the graph of $u = 20 - v^{2}$ and $v = 20 -u^{2}$ on $(u, y)$-plane, which gives two parabolas that are reflections each other with respect to the line $u = v$. And, you have another guess: The other solutions do not satisfy $u\geq 0$ and $v\geq 0$, which should happen since $u = x^{2}$ and $v = y^{2}$. (Note that there are 3 more intersection points) So we can say confidently that the only solution is $(u, v) = (2, 2)$, but this is not a proof. There's another solution that you can guess that corresponds to another intersection point with $u = v$ other than $(4, 4)$. If you set $u = v$, then you obtain $u = -5, 4$ by solving a quadratic equation.
At last, to prove that this is the only solution, note that the degree 4 polynomial should be divisible by $(v-4)$ and $(v+5)$ since $4, -5$ are roots of it. We have
$$
v^{4} - 40v^{2} +v + 380 = (v-4)(v+5)(v^{2} - v - 19)
$$
and $v^{2} - v - 19  = 0 \Leftrightarrow v = \frac{1 \pm \sqrt{77}}{2}$, where both of them fail to satisfy $(u \geq 0) \wedge (v \geq 0)$. So the only answer is $(u, v) = (2, 2) \Leftrightarrow (x, y) = (\pm 2, \pm 2)$.
A: You have the following problem: $$\begin{bmatrix}x^2+y^4=20\\ x^4+y^2=20\end{bmatrix}$$
In the first equation, you can isolate $x$ to get $x=\sqrt{20-y^4}$ and $\:x=-\sqrt{20-y^4}$. You then substitute the $x$ values into the second equation to get:
$$\left(\sqrt{20-y^4} \right)^4 +y^2=20, \left(-\sqrt{20-y^4} \right)^4 +y^2=20$$
Simplifying the equations will result in $y=2, y=-2$. You can then substitute the solution into the first equation and you will get $x=2,-2$ depending on which value of $y$ you substitute. Therefore, the final solutions are:
$$\begin{pmatrix}x=2,\:&y=2\\ x=-2,\:&y=2\\ x=2,\:&y=-2\\ x=-2,\:&y=-2\end{pmatrix}$$
