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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?

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    $\begingroup$ @AlbusDumbledore, thank you for the response! I would like a solution without guessing, though. Can I ask what do you mean by "due to symmetry"? What is symmetric in the system? $\endgroup$ Commented Feb 1, 2021 at 13:49
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    $\begingroup$ What on earth is with all of the downvotes on the answers here? Seems like someone is serially downvoting (of sorts)... $\endgroup$ Commented Feb 1, 2021 at 13:54
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    $\begingroup$ @CameronWilliams with you on that $\endgroup$ Commented Feb 1, 2021 at 13:56
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    $\begingroup$ I flagged for moderator attention since this is very negative behavior. $\endgroup$ Commented Feb 1, 2021 at 13:56
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    $\begingroup$ @CameronWilliams good,it is definetly intentional,because after 2 seconds or so after the answer being posted there is a downvote,I cant fathom the downvoter to read an anwer within 2 seconds unless the downvoter is a superhuman $\endgroup$ Commented Feb 1, 2021 at 13:58

5 Answers 5

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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 :)

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    $\begingroup$ Now this is a good answer(upvoted with all force i could muster!) $\endgroup$ Commented Feb 1, 2021 at 14:03
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    $\begingroup$ Thank you all very much! +64 is a huge amount ;) (= $\endgroup$ Commented Feb 1, 2021 at 14:16
  • $\begingroup$ ultraedge you mean +65? $\endgroup$ Commented Feb 1, 2021 at 14:17
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    $\begingroup$ Someone downvoted! Plus some other reps from other posts. $\endgroup$ Commented Feb 1, 2021 at 14:18
  • $\begingroup$ But no one has downvoted this post $\endgroup$ Commented Feb 1, 2021 at 14:19
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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".

enter image description here

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    $\begingroup$ This downvote is undeserved. $\endgroup$
    – user65203
    Commented Feb 1, 2021 at 13:54
  • $\begingroup$ [+1] from me. Thank you! $\endgroup$ Commented Feb 1, 2021 at 13:54
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    $\begingroup$ The fundamental theorem of algebra says us that the equation has at most four solutions IN $v$, and you only suggested TWO possible $v$'s, so there might be more (actually there are) $\endgroup$
    – Seewoo Lee
    Commented Feb 1, 2021 at 14:00
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    $\begingroup$ "...you get a polynomial of degree four": degree eight, surely? It is of degree four in $x^2$, but four solutions for $x^2$ means up to eight solutions for $x$. So I think this answer is wrong. $\endgroup$
    – TonyK
    Commented Feb 1, 2021 at 14:03
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    $\begingroup$ @IDoktorova: I made a mistake; I edited my answer. $\endgroup$
    – user9464
    Commented Feb 1, 2021 at 16:06
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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.

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    $\begingroup$ I dont know who it is but there is one downvoter who is roaming here to downvote each and every post including the question! $\endgroup$ Commented Feb 1, 2021 at 13:55
  • $\begingroup$ Can you explain to me how do we get from the first to the second line? $\endgroup$ Commented Feb 1, 2021 at 13:58
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    $\begingroup$ @IDoktorova expand the second equation. It'll be clear from there. It's just a clever form of $0$. $\endgroup$ Commented Feb 1, 2021 at 13:58
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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 $(u, v) = (2, 2) \Leftrightarrow (x, y) = (\pm 2, \pm 2)$.

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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:

$$\begin{pmatrix}x=2,\:&y=2\\ x=-2,\:&y=2\\ x=2,\:&y=-2\\ x=-2,\:&y=-2\end{pmatrix}$$

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