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Here

http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html

it is claimed that a solution of the diophantine equation $$ax^2+bxy+cy^2=1\ \ \ \ (1)$$ is always a convergent of a root of $ax^2+bx+c$.

I found out that a solution of $(1)$ can exist even if $ax^2+bx+c$ has no real root, but in the example $$3x^2+9xy+7y^2=1$$ , a solution is $(3/-2)$ and $-\frac{3}{2}$ is the real part of both roots of $3x^2+9x+7$.

Is this a coincidence ?

Moreover, if the discriminant is $0$, for example $$4x^2+20xy+25y^2=1$$ has solution $(8/-3)$ , but $4x^2+20x+25=0$ has solution $-\frac{5}{2}$ , which does not have convergent $-\frac{8}{3}$.

And the case $a=0$ seems to work only if $b>2$, but this case is not very interesting anyway.

My work so far to prove the claim :

If $(s/t)$ is a solution of $(1)$ with $t\ne 0$ and $\alpha$ is a root of $ax^2+bx+c$, then we have

$$a(\frac{s}{t})^2+b\frac{s}{t}+c=\frac{1}{t^2}$$

and $$a\alpha^2+b\alpha+c=0$$

This implies $$a(\frac{s}{t}-\alpha)(\frac{s}{t}+\alpha)+b(\frac{s}{t}-\alpha)=\frac{1}{t^2}$$ implying $$\frac{s}{t}-\alpha=\frac{1}{t^2(a(\frac{s}{t}+\alpha)+b)}$$

This looks promising because we only have to estimate $a(\frac{s}{t}+\alpha)+b$.

Who has an idea how to complete the proof and find out the conditions when the solution of $(1)$ must be a convergent of one of the roots ?

Peter
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  • The link given by You implies a different approach. Consideration by the General formula involves the use of several formulas. http://math.stackexchange.com/questions/1513733/solving-a-diophantine-equation-of-the-form-x2-ay2-byz-cz2-with-the-co/1514030#1514030 Solvability of an equation implies finding out some equivalent to the solvability of the Pell equation. – individ Feb 22 '17 at 10:53

1 Answers1

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In the direction you want, all that is required is a theorem of Legendre, that a rational number $p/q$ with $\gcd(p,q) = 1$ and $q > 0$ is a convergent to some real $\alpha$ if $$ \left| \alpha - \frac{p}{q} \right| < \frac{1}{2 q^2} $$ For example, I wrote out the case when $x,y > 0,$ $$ A x^2 + B xy + C y^2 = 1 $$ and $$ A > 0, B > 1, C < 0, $$ so that $$ \Delta = B^2 - 4 A C > B^2. $$ Taking $r = x/y$ it became $$ \left( r - \left( \frac{-B + \sqrt \Delta}{2} \right) \right)\left( r + \left( \frac{B + \sqrt \Delta}{2} \right)\right) = \frac{1}{A y^2}, $$ $$ r + \left( \frac{B + \sqrt \Delta}{2} \right) > \left( \frac{B + \sqrt \Delta}{2} \right) $$ $$ 0 < \left( r - \left( \frac{-B + \sqrt \Delta}{2} \right) \right) < \frac{1}{A y^2} \frac{2A}{B + \sqrt \Delta} = \frac{2}{ y^2 (B + \sqrt \Delta)} < \frac{1}{2 y^2} $$

The first book is Burger, the second book is Khinchin.

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Will Jagy
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