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Steps to solve this system of equations

During the flight from Moscow to Yerevan my neighbor gave me the following problem:

Solve the system:

$$\left\{\begin{array}{c}x^2+y=7 \\ y^2+x=11. \end{array}\right.$$

It is easy to find 1 of the 4 solutions. Is there a beautiful way to find the other three?

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marked as duplicate by Gerry Myerson, Aryabhata, Willie Wong May 15 '12 at 15:23

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4  
Doesn't seem so - press 'Exact forms' to see analytical solution. Back in Russia, I was told of a similar problem $$ \begin{cases} x^2+y &= 35, \\ x+y^2 &= 105 \end{cases} $$ with a nice solution $x=5,y = 10$, but others appeared to be hard to find Anyway, I'm interested in ways to solve these problems with using Cardano's formulae –  Ilya May 14 '12 at 9:14
    
Are these supposed to be integer solutions? The system itself looks like it describes a circle. –  chris May 14 '12 at 9:18
    
Ilya, i hope for some beautiful trick to solve this without Cardano's formulae. –  Roah May 14 '12 at 9:20
    
chris, there is one integer solution and three irrational solutions. I wonder whether there is a beautiful way to find these three solutions. –  Roah May 14 '12 at 9:21
2  
You get an irreducible cubic. The solutions to an irreducible cubic are guaranteed to be messy. There can't be a beautiful way to find ugly solutions. –  Gerry Myerson May 14 '12 at 13:06
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2 Answers 2

up vote 11 down vote accepted

Every solution of the given system

$$\left\{ \begin{array}{c} x^{2}+y=7 \\ y^{2}+x=11 \end{array} \right. \tag{0}$$

is a solution of

$$\left\{ \begin{array}{c} \left( y-3\right) \left( y^{3}+3y^{2}-13y-38\right) =0 \\ x^{2}=121-22y^{2}+y^{4}. \end{array}\tag{1} \right. $$

The same applies to the system

$$\left\{ \begin{array}{c} y^{2}=49-14x^{2}+x^{4} \\ \left( x-2\right) \left( x^{3}+2x^{2}-10x-19\right) =0. \end{array}\tag{2} \right. $$

The integral solution of $(0)$ is $\left( x_{0},y_{0}\right) =\left( 2,3\right) $. Simple ways to find the remaining solutions are only possible in particular cases, as far as I know. The standard way to solve a cubic equation such as

$$y^{3}+3y^{2}-13y-38=0\tag{3}$$

is to make the change of variables $$y=s-\dfrac{3}{3\cdot 1}=s-1\tag{3a}$$ to get the reduced cubic equation

$$s^{3}-16s-23=0.\tag{4}$$

If the discriminant $q^{2}+\frac{4p^{3}}{27}$ of an equation of the form $s^3+px+q=0$ is negative, its three solutions are real numbers. In this case we have $q^{2}+\frac{4p^{3}}{27}=23^{2}-\frac{4\times 16^{3}}{27}<0$ and the solutions of $(4)$ can be written in the trigonometric form$^{1}$

$$s_{k}=2\sqrt{\frac{16}{3}}\cos \left( \frac{1}{3}\arccos \left( \frac{23}{2}\sqrt{\frac{27}{16^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right),\tag{5}$$

with $k=1,2,3$. So $$y_{k}=s_{k}-1\tag{6}$$ and

$$x_{k}=11-y_{k}^2.\tag{7}$$

For $k=1$, we get $\left( x_{1},y_{1}\right) \approx \left( -1.8479,3.5844\right) $. And similarly for $k=2$ and $k=3$.

--

$^{1}$A deduction can be found in this Portuguese post of mine.

Added. If $\Delta =q^{2}+\frac{4p^{3}}{27}<0$ the three real solutions of the following reduced cubic equation

$$t^{3}+pt+q=0\tag{A}$$

are given by

$$t_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}\sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right) \tag{B},$$

with $k=1,2,3$.

PS. I do not find trigonometric functions nor radicals ugly. But this is just an opinion.

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1  
Thanks for the detailed solution, but this is not exactly what i am searching for. This is a standard trigonometric substituion. I am in a quest for the solution "from the book" particularly tailored for this problem. I expect an answer in a much simplier form. If it will not be found in two weeks i will accept you answer as a final one. –  Roah May 14 '12 at 11:58
    
@Roah Thanks! Do you mean an answer not in terms of radicals nor in terms of trigonometric functions? –  Américo Tavares May 14 '12 at 12:01
    
Given how ugly the answer is, do you really expect there's a nicer way to get to it? –  Gerry Myerson May 14 '12 at 13:03
1  
Yes (sorry, should have made that clear). –  Gerry Myerson May 14 '12 at 13:15
1  
@AméricoTavares: I'm sorry to hear about downvoters again. Please take my upvote for a nice answer - maybe it will balance away that downvote! –  Ilya May 14 '12 at 16:03
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This is maybe a half solution, maybe less. I tried several things and maybe these ideas inspire someone to write a complete solution.

Note that the intersection points lie on the circle of the form

$$\left(x+\frac12\right)^2+\left(y+\frac12\right)^2=\frac{37}{2}$$

This follows from a general theorem on the intersections of two parabolas with orthogonal axes of symmetry.

Now there are so many theorems on cyclic quadrilaterals and even some statements on a parabola passing through the four corners. Then the given point $(2,3)$ should give some additional information. But I couldn't find a concise way of finishing this.

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3  
You can also get this equation by adding the two given ones and completing the squares. No beautiful conic theorems needed. –  Ross Millikan May 14 '12 at 13:27
    
Yeah sure, the formula is not hard to find. I just think that this approach might eventually give a really nice representation of the four points. In the end there are explicit formulas for all four points just depending on three angles. Then there are planty of relations between many of the involved line segments. And also the symmetry axis of the parabola is somehow related to some of those segments. I think there is even a transformation which breaks down to reflection at some circle which gives some sort of normal form... I just couldn't put the pieces together. –  Simon Markett May 14 '12 at 13:38
    
Since this answer already got six upvotes even though it doesn't show anything, I am trying to improve it. If anyone has any input please tell me! –  Simon Markett May 14 '12 at 18:04
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