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Let $n_{1}$ and $n_{2}$ be positive rational numbers such that $n_{1}<n_{2}$. Let $a=3n_{1}$, $b=-3n_{1}^2$, $c=n_{1}^3-n_{2}^3$.

Can $$\frac{-b+\sqrt{b^2-4ac}}{2a}$$ be a rational number?

In my problem, see Parcly Taxel answer, I want $h_{w_1},h_{w_2}$ and $h$ all to be positive rational numbers with $(h_{w_1}<h_{w_2})$. Is that possible?

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Hussain-Alqatari
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2 Answers2

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EDIT: This post has missed one important factor that $n_1 < n_2$, what means that this post does not answer the question.

I assume your talking about quadratic equations and the quadratic formula. Then you have the input $3n_1x^2 - 3n_1x + n_1^3 - n_2^3 = 0$, then the quadratic formula $\frac{-b + \sqrt{b^2 -4ac}}{2a}$. If we insert all the values, we get the following term:

$$\frac{3n_1 + \sqrt{9n_1^2 -4(3n_1)(n_1^3 - n_2^3)}}{6n_1} \in \mathbb{N}$$ $$n_1 \ne 0$$

Since the question only asks "Can it be rational?", we just need to find a pair that makes the equation rational: Since $n_1$ is rational, we can ignore everything outside of the root, giving us:

$$\sqrt{9n_1^2 -4(3n_1)(n_1^3 - n_2^3)} \in \mathbb{N}$$ $$n_1 \ne 0$$

Reform the equation

$$\sqrt{9n_1^2 -4(3n_1)(n_1^3 - n_2^3)} = \sqrt{9n_1^2 - 12n_1^4 + 12n_1n_2^3}$$

Now the goal is to show that $\sqrt{9n_1^2 - 12n_1^4 + 12n_1n_2^3}$ is a rational number, which reformed gives $\sqrt3\sqrt{3n_1^2 - 4n_1^4 + 4n_1n_2^3}$. This means that the right root must be of the form $3 \cdot x^2$, so the output makes sense.

Someone
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    I know that already .. it is true. But we need $n_{1}<n_{2}$. Is it possible? – Hussain-Alqatari Aug 07 '19 at 12:22
  • Oh, my bad! I have overseen that one! Wait let me fix the post - give me one hour before voting anything, please! – Someone Aug 07 '19 at 12:43
  • I've wasted my entire day trying to answer your question hunting for reputation. Now I've wasted another hour to try to fix my mistake... @Hussain-Alqatari, sadly I couldn't find an answer, but I am pretty sure that there is no answer for the conditions you gave. I've managed to find out that both n's have to be divisible by 3, maybe that helps. I'm sorry. Looks like I'm more stupid than I thought :) – Someone Aug 07 '19 at 13:51
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    No stupid person can post what you have posted. Appreciated ,, thanks! :) – Hussain-Alqatari Aug 07 '19 at 13:54
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We look at a more general case where we only require $a,b,c$ and $(-b+\sqrt{b^2-4ac})/(2a)$ to be rational.

A necessary condition is $$y^2 = b^2-4ac = 3n_1(4n_2^3-n_1^3)$$ for some rational $y$. Since we cannot have $n_1=0$ due to the denominator of $2a$, we may assume that $n_1\neq 0$.

Multiplying by $144/n_1^4$ and setting $n_2 = \dfrac{n_1X}{12},y=\dfrac{n_1^2Y}{12}$, we get an Elliptic Curve $E$ over the rationals $$ E: Y^2=X^3-432 $$ This tells us that every rational solution $(a,b,c,y)$ to our original equation must map to some rational point $(X,Y)$ on $E$. Hence we can start from $E$ and work backwards.

Now the critical part is this Elliptic Curve $E$ is well-known to only have the solutions $(X,Y) = (12,\pm 36)$ over the rationals. One reference from here.

This means that the only possible solutions of $(a,b,c,y)$ must satisfy $$ \begin{align*} (12,\pm 36) &= (X,Y) = \left(\frac{12n_2}{n_1}, \frac{12y}{n_1^2}\right)\\ (1,\pm 3) &= \left(\frac{n_2}{n_1},\frac{y}{n_1^2}\right) \end{align*} $$ This shows that we must have $n_2/n_1=1$.

Now going back to the original question, since we require $n_1<n_2$ this is impossible.

Yong Hao Ng
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