# 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?

• @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? – Medi Feb 1 at 13:49
• What on earth is with all of the downvotes on the answers here? Seems like someone is serially downvoting (of sorts)... – Cameron Williams Feb 1 at 13:54
• @CameronWilliams with you on that – Albus Dumbledore Feb 1 at 13:56
• I flagged for moderator attention since this is very negative behavior. – Cameron Williams Feb 1 at 13:56
• @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 – Albus Dumbledore Feb 1 at 13:58

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 :)

• Now this is a good answer(upvoted with all force i could muster!) – Albus Dumbledore Feb 1 at 14:03
• Thank you all very much! +64 is a huge amount ;) (= – ultralegend5385 Feb 1 at 14:16
• ultraedge you mean +65? – Albus Dumbledore Feb 1 at 14:17
• Someone downvoted! Plus some other reps from other posts. – ultralegend5385 Feb 1 at 14:18
• But no one has downvoted this post – Albus Dumbledore Feb 1 at 14:19

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". • This downvote is undeserved. – Yves Daoust Feb 1 at 13:54
• [+1] from me. Thank you! – Medi Feb 1 at 13:54
• 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) – Seewoo Lee Feb 1 at 14:00
• "...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. – TonyK Feb 1 at 14:03
• @IDoktorova: I made a mistake; I edited my answer. – user9464 Feb 1 at 16:06

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.

• I dont know who it is but there is one downvoter who is roaming here to downvote each and every post including the question! – Albus Dumbledore Feb 1 at 13:55
• Can you explain to me how do we get from the first to the second line? – Medi Feb 1 at 13:58
• @IDoktorova expand the second equation. It'll be clear from there. It's just a clever form of $0$. – Cameron Williams Feb 1 at 13:58

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 $$(u, v) = (2, 2) \Leftrightarrow (x, y) = (\pm 2, \pm 2)$$.

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:

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