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This is the system of equations: $$\sqrt { x } +y=7$$ $$\sqrt { y } +x=11$$

Its pretty visible that the solution is $(x,y)=(9,4)$

For this, I put $x={ p }^{ 2 }$ and $y={ q }^{ 2 }$. Then I subtracted one equation from the another such that I got $4$ on RHS and factorized LHS to get two factors in terms of $p$ and $q$.

Then $4$ can be represented as $2*2$, $4*1$ or $1*4$. Comparing the two factors on both sides, I got the solution.

As you can see, the major drawback here is that I assumed this system has only integral solutions and then went further. Is there any way I can prove that this system indeed has only integral solutions or is there any other elegant way to solve this question?

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Maybe not too elegant, but certainly worth trying: Graphing –  imranfat Dec 11 '13 at 3:53
    
What is the maximum number of solutions this type of system can have? (Because if you find all of the solutions, and they're all integer solutions, you've won.) –  John Dec 11 '13 at 4:00
4  
The current answers don't have enough detail???? –  Igor Rivin Dec 13 '13 at 18:09

4 Answers 4

up vote 12 down vote accepted
+100

You want \begin{cases} x=(7-y)^2\\ y=(11-x)^2\\ 0\le x\le 11\\ 0\le y\le 7 \end{cases}

The equation becomes $$ x=49-14(121-22x+x^2)+(121-22x+x^2)^2 $$ which reduces to $$ (x-9)(x^3-35x^2+397x-1444)=0 $$ (courtesy of WolframAlpha). The polynomial $f(x)=x^3-35x^2+397x-1444$ has at least one real root. It has indeed three. One of them satisfies the condition $0\le x\le 11$ and it's approximately $7.87$. With this value of $x$ we get $y\approx 9.79$ that doesn't satisfy $0\le y\le 7$.

One can be more precise: call $\alpha$ the least root of $f$. Then $7<\alpha<8$, so that $3<11-\alpha<4$ and so $9<(11-\alpha)^2<16$, which shows that the limitation on $y$ is not fulfilled.

How can you know this? Compute

  • $f(7)=-37$
  • $f(8)=4$
  • $f(12)=8$
  • $f(13)=-1$
  • $f(15)=11$

Thus you know that the three roots of $f$ are $7<\alpha<8$, $12<\beta<13$ and $13<\gamma<15$.

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Solve the first equation for $x,$ the second for $y,$ to get

$$ x = (y-7)^2.$$

$$ y = (x-11)^2.$$

Substitute the value for $x$ into the second equation, to get:

$$y = ((y-7)^2 -11)^2 = (y^2-14 y + 38)^2.$$ This is a quartic equation, but dividing through by $y-4,$ you get an irreducible cubic $y^3-24y^2+176 y - 361.$ Now, being cubic, it has at least one real solution, which would indicate that there is at least one more real pair $(x, y).$

EDIT Mathematica seems to think that the only solution is $9, 4.$

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Just to back up. $y_2 \approx 3.4156$, $y_3 \approx 9.8051$, $y_4 \approx 10.7793$. –  Ray Dec 13 '13 at 18:09
    
Same comment as on rajb245's answer. These don't seem to actually satisfy the original equation unless I'm doing something very wrong with my calculator. Are you sure they're not extraneous? –  Logan Maingi Dec 20 '13 at 4:37
    
What does "extraneous" mean? –  Igor Rivin Dec 20 '13 at 4:59
    
@IgorRivin Extraneous solutions are values which arise when one performs noninvertible operations on a set of equations which leads to a necessary, but not sufficient, new set of equations. Those values which do not solve the original set of equations are extraneous. For example, $x=2$ obviously only has one solution, but if one squares both sites the resulting equation $x^2 = 4$ has another solution, $x=-2$, which is "extraneous". –  Logan Maingi Dec 20 '13 at 5:07
    
@IgorRivin, yes, Logan is right. The values do not solve the original equations, and this is because we squared the original equations, which of course clobbers the $\pm$ sign of all squared quantities. The correct approach, I think, is to follow this procedure, try all the four candidate solutions in the original equations, and conclude that only one of them actually works. It consists of integers. –  rajb245 Dec 21 '13 at 16:24

If you move things around and square the equations, you have the following two starting equations. $$x=(7-y)^2$$ $$y=(11-x)^2$$

If you plug one of these into the other and factor out a term, you get $$ (x-9)(x^3-35x^2+397x-1444)=0. $$

This is a quartic equation, with four solutions in general. There is a linear term, $x-9$, and the remaining cubic term. If $x=9$, the linear term goes to zero and the equation is solved. If you plug $x=9$ back into the original equations, you get $y=4$, so this represents the original solution you found. To find the other solutions, we only have to focus on the roots of the remaining cubic term. So we're left with finding the roots of this equation

$$ x^3-35x^2+397x-1444=0. $$

All cubics have three roots, if you count complex roots and double roots. The possibilities go like this for any cubic:

There are three real, distinct roots

Three roots

There are three real roots, but two of them are merged into a double root

enter image description here

There is one real root and a pair of complex conjugate roots

enter image description here

You can follow along with this procedure from Wikipedia to calculate the exact roots of the equation. Or you can use a computer solver for the roots that does a similar procedure internally:

enter image description here

This shows you that this cubic is in the first family with three real, distinct roots, with $x \approx 7.8687, 12.848, 14.283$ and corresponding $y\approx 9.80504, 3.4151, 10.7781$.

Edit

As a user pointed out, the values we just found should be plugged into the original equations to see if they actually solve the equation, or if they are extraneous. It turns out that these three solutions are indeed extraneous, meaning that they are spurious results of squaring the original equations. This leaves the only solution we've found as $(9,4)$. It is necessary that the solutions are a subset of the four presented, because the system is fundamentally fourth order (two quadratic equations), so it can have at most four solutions. The only one of the necessary solutions that is sufficient is $(9,4)$, so this is the unique solution. It consists of two integers, so we have proof that this equation only has integer solutions.

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Am I missing something here? I tried plugging the values back into the equations in the OP, and it isn't even close to being solved. It seems that these solutions are extraneous which do not solve the original system of equations. –  Logan Maingi Dec 20 '13 at 4:36
    
Hm, I didn't do the basic check of plugging back into the original equations....and I'm getting that the presented values don't solve the original system. I'll get back to you... –  rajb245 Dec 21 '13 at 15:59
    
Edited to fix the erroneous logic. Great catch, Logan! –  rajb245 Dec 21 '13 at 16:22
    
@egreg honestly had the best answer. He did says that of the four possible solutions, only one satisfies all the conditions. –  rajb245 Dec 21 '13 at 16:29

Second equation implies $y$ is a perfect square, first one implies it's at most $7$. There are only 3 options - $y=0$, $y=1$, $y=4$. The first two don't work (they lead to $x=49$ and $x=36$ respectively), so $y=4$, $x=9$ is the only integral solution.

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For completeness one should add, that $x,y \ge 0$ for to have the solutions non-imaginary. After that the constants on the rhs give upper bounds for the compositions on the lhs. After that the integer values for x and y , which have also integer roots, can be enumerated to arrive at your solution ... (I put this under your answer as an comment because I like the conciseness of your exposition) –  Gottfried Helms Dec 20 '13 at 0:00

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