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is there a method to find all the solutions to the following set of irrational equations,



NOTE: $(4-1)=(2-1)(2+1)=3$ and $(4+1)=(2^2+1^2)=(3^2-2^2)=(3-2)(3+2)=5$

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The title does not match the question and neither of which match the tag. The note seems completely irrelevant and pointless unless you're trying to lead people in a specific direction. – Mike Aug 21 '12 at 14:28
Maybe then just saying "Yes, there is" is the right answer, since it applies to both questions. – Zarrax Aug 21 '12 at 14:32
the note indicates that $\sqrt{x}=2,y=1$ is one of the solution(s). – Rajesh K Singh Aug 21 '12 at 14:34
@RajeshKSingh: If the equation in the title is what you got by manipulating one equation and then substituting into the other equation, you should right it explicitly as a solution attempt. – user3533 Aug 21 '12 at 14:35
@Mike And, yet, the question has 3 upvotes and is favorited by 1 user. – Graphth Aug 21 '12 at 16:40
up vote 1 down vote accepted

If you make the change of variables $$\begin{equation*} u=\sqrt{x}\geq 0,\qquad v=\sqrt{y}\geq 0, \end{equation*}$$

then you need to solve $$\begin{equation*} \left\{ \begin{array}{c} u+v^{2}=3 \\ u^{2}+v=5 \end{array} \right. \end{equation*}$$

or $$\begin{eqnarray*} &&\left\{ \begin{array}{c} u=3-v^{2} \\ v^{4}-6v^{2}+v+4=0 \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} u=3-v^{2} \\ \left( v-1\right) \left( v^{3}+v^{2}-5v-4\right) =0. \end{array} \right. \end{eqnarray*}$$

So $u=2,v=1$ is a solution, i.e. $x=4,y=1$ in the original variables. And since $v^{3}+v^{2}-5v-4=0$ has 3 real solutions, one positive $v_{1}\approx 2.164\,2\gt \sqrt{3}$ and two negative $v_{2}\approx -2.391\,4,v_{3}\approx -0.772\,87$, there are no other solutions because $u=3-v_{1}^{2}<0$.

share|cite|improve this answer there a method to find all the solutions[?]

Yes, there is: draw the graphs of the functions $x\mapsto3-\sqrt{x}$ and $x\mapsto(5-x)^2$ on $[0,5]$, these intersect at $(4,1)$ and there only hence the unique solution of your system is $(x,y)=(4,1)$.

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For the question in the title, you can move the $\sqrt y$ to the other side, square, and get a quartic which yields to the rational root theorem.

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Which says nothing about possible other roots. – Did Aug 22 '12 at 11:28




i.e. $\sqrt{y}=1$, and $x=4$ is a solution

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You should probably edit your question and add this there to show how everything relates and show your effort. For future reference, diophantine equations involve finding rational or integer solutions to equations. I'll change my vote on the question and retag it to something more appropriate. – Mike Aug 21 '12 at 15:20

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