# Rational solutions for quadratic equations in two variables [closed]

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.

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

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

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

... 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 $$b$$.

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