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I am looking for a generalisation of the solution of $x,y$ wich are rational numbers,they could be infinite,how can i find such solutions,integer solutions are obvious

I have found that

$\Delta=-11,u=1,d=-11$

also

$(x+\frac{3+\sqrt{-11}y}{2})(x+\frac{3-\sqrt{-11}y}{2})=4$

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  • $\begingroup$ By obvious you mean Pell equations? $\endgroup$ – chubakueno Apr 5 '14 at 16:45
  • $\begingroup$ yes ,i am trying to say in this case i am not interested in integer solutions,but rational $\endgroup$ – Jonas Kgomo Apr 5 '14 at 16:52
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We have the simple solution $x=2$, $y=0$. Now take the line though $(2,0)$ with slope $m$, and find where it meets the ellipse. That will give a rational parametrization of the ellipse.

Details: The line has equation $y=m(x-2)$. Substitute. We get $x^2+3xm(x-2)+5m^2(x-2)^2=16$. This simplifies to $(1+3m+5m^2)x^2-(6m+20m^2)x+20m^2-4=0$. The product of the roots is $\frac{20m^2-4}{1+3m+5m^2}$. But one of the roots is $2$, so the other is $\frac{10m^2-2}{1+3m+5m^2}$. Now we can compute the corresponding $y$. Any rational $m$ will give us a rational solution of the original equation.

Remark: The procedure was general. In particular, let $ax^2+bxy+cy^2=d$ be an ellipse with $a,b,c,d$ rational. If we know a rational point on the ellipse, then we can find a parametric expression for all rational points. The most important special case is the circle.

And the basic idea generalizes, importantly, to elliptic curves.

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  • $\begingroup$ Your remark is what makes your answer better than mine. But I would like to know: Is there an efficient method to get a rational point? Can we know when does it exist, or some bound on the denominator and numerator? $\endgroup$ – chubakueno Apr 5 '14 at 17:52
  • $\begingroup$ Yes, there is a huge amount of theory of ternary quadratic forms (it is useful to homogenize, and consider $ax^2+bxy+cy^2=4z^2$.) The details are complicated, they go back to Legendre and Gauss, and are a large chapter in the history of number theory. $\endgroup$ – André Nicolas Apr 5 '14 at 17:59
  • $\begingroup$ i am actually interested in that,more about binary and not tenary ,can i say that solution is for both rational and integers $\endgroup$ – Jonas Kgomo Apr 5 '14 at 18:07
  • $\begingroup$ The integer solutions of the ternary (homogenized) form are in a simple way connected to the rational solutions of the original equation. $\endgroup$ – André Nicolas Apr 5 '14 at 18:10
  • $\begingroup$ i would appreciate if both of you can help me with this as well math.stackexchange.com/questions/739752/… $\endgroup$ – Jonas Kgomo Apr 5 '14 at 18:14
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This is a variation of Hecke's answer $$(2x+3y)^2+11y^2=4^2$$ $$\left(\frac{2x}y+3\right)^2+11=\left(\frac4y\right)^2$$ $$11=\left(\frac4y\right)^2-\left(\frac{2x}y+3\right)^2$$ $$11=\left(\frac{4-2x}y-3\right)\left(\frac{4+2x}y+3\right)$$ $$\frac{4-2x}y-3=t, \frac{4+2x}y+3=\frac{11}t$$ $$\implies x = -\frac{2(t^2+6t-11)}{t^2+11}, y = \frac{8t}{t^2+11}, t\neq0$$


My comment contained an imprecision. This works for every equation that can be transformed into the form: $$(ax+by)^2+cy^2=\left(\frac pq\right)^2$$ Where $a,b,c,p,q\in\mathbb Z, q\neq 0$

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  • $\begingroup$ this is excellent,it seems like it will absolutely work for any binary equation $\endgroup$ – Jonas Kgomo Apr 5 '14 at 18:11
  • $\begingroup$ is $t$ an integer $\endgroup$ – Jonas Kgomo Apr 5 '14 at 18:16
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    $\begingroup$ @Jonas12: $t$ is any rational number. $\endgroup$ – TonyK Apr 5 '14 at 18:18
  • $\begingroup$ @Jonas12 Actually we were quite lucky :) .When we completed the square in the LHS multiplying by $4$, $4^2$ resulted to be a square. If you change that $4$ by $5$(or any non-square number) in your equation we would have a problem. In particular, this works with the general equation $ax^2+bxy+cy^2=d$, when $d$ is a perfect square. André Nicolas solutions helps us here if it is not, if we can find a rational solution. $\endgroup$ – chubakueno Apr 5 '14 at 18:23
  • $\begingroup$ the manner of the solution proposed by the proffesor,is in the manner of the solution on page 47-48 for the problem $x^2+3xy-5y^2=65$ i am wondering if this approach and @Andre Nicolas solutions will be acceptable $\endgroup$ – Jonas Kgomo Apr 5 '14 at 18:42

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