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Consider the following quadratic equation where $a, b > 0$ are integers:

  • $x_2^2 = x_1^2 - ax_1 + b^2$

I am looking for the rational solutions of the above equation satisfying the following conditions:

  • $0 < x_1 < \frac{a}{2}$

  • $\sqrt{b^2 - \frac{a^2}{4}} < x_2 < b$

  • $2b > a$

Now consider the following conjectures:

  • Conjecture 1: There are infinitely many rational values of $x_1, x_2$ satisfying the above equation for each possible positive integer values of $a, b$.

  • Conjecture 2: If the Conjecture 1 is true, for a given integers $a, b$, the rational values of $x_1, x_2$ satisfying the above equation form a dense set.

Questions:

  • Are the above conjectures true ? Are there any similar known conjectures or results ?

  • If the above conjectures are false, what are the conditions (necessary and/or sufficient) on the integer values of $a, b$ such that the above conjectures are true.

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  • $\begingroup$ $x_1=0, x_2=b$ surely works, integer values too. $\endgroup$
    – Macavity
    Oct 12, 2021 at 5:56
  • $\begingroup$ @Macavity I have added more details and boundary conditions. $\endgroup$ Oct 12, 2021 at 14:56

1 Answer 1

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Write your original equation as:

$$\left(x_1-\frac{a}{2}\right)^2-x_2^2=\frac{a^2}{4}-b^2$$

$$\left(x_1-\frac{a}{2}-x_2\right)\left(x_1-\frac{a}{2}+x_2\right)=\frac{a^2}{4}-b^2$$

... so you could equate, for example $x_1-x_2-\frac{a}{2}=1$, $x_1+x_2-\frac{a}{2}=\frac{a^2}{4}-b^2$ and get a rational solution for $x_1$ and $x_2$.

So the answer to your question is yes, there is an infinite number of solutions for each choice of integers $a$ and $b$.

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  • $\begingroup$ I have added more details and boundary conditions. $\endgroup$ Oct 12, 2021 at 14:55
  • $\begingroup$ @ShivaKintali Then you should also have $2b>a$ for the square root to make sense. $\endgroup$
    – Momo
    Oct 12, 2021 at 16:08
  • $\begingroup$ Yes. I have added it now. $\endgroup$ Oct 12, 2021 at 17:38

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