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Description: Given the following parametric cubic polynomials ${E}^{3} - 15\, {\beta}_{\pm}\, {E}^{2} - 3 \left({71\, {\beta}_{\pm}^{2} + 352\, {\beta}_{\mp}}\right) E + 135\, {\beta}_{\pm} \left({5\, {\beta}_{\pm}^{2} - 32\, {\beta}_{\mp}}\right)$ where ${\beta}_{\pm} = 1 \pm \eta$, how do I find all possible rational values of $\eta$ such that this polynomial is reducible.

Background: I have found 61 rational values of $\eta$ that factor these polynomials into a rational linear $E - p/q$ for integer $p$ and $q$ relatively prime and $q \ne 0$ times an irreducible quadratic of the form ${E}^{2} + a\, E + b$. I do not know if there are an infinite number of such values and if not do I have the complete set.

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  • $\begingroup$ I updated my posting and simplified the description for clarity. Simply put I am looking for all the cases where these parametric cubics are reducible over rationals. $\endgroup$
    – Lorenz H Menke
    Commented Mar 16, 2015 at 23:16
  • $\begingroup$ The curve is rational. $\endgroup$
    – Felipe Voloch
    Commented Mar 17, 2015 at 1:32
  • $\begingroup$ Yes I am looking for the rational points. $\endgroup$
    – Lorenz H Menke
    Commented Mar 17, 2015 at 1:36
  • $\begingroup$ @LorenzMenke I said the curve is rational, as in parametrizable by rational functions. $\endgroup$
    – Felipe Voloch
    Commented Mar 17, 2015 at 1:42

1 Answer 1

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I just reopened the question so I can write an answer, although my comment should have sufficed but I may not be getting my message across.

Anyway, let's use some sane variables ($x=\beta_{\pm},y = E$), so the polynomial is

$$y^3-15xy^2-3(71x^2+352x)y+135x(5x^2-32x)$$

Let's now replace $y$ by $xz$ and get

$$(z^3 - 15z^2 - 213z + 675)x^3 + (-1056z - 4320)x^2.$$

So you can pick any rational $z$ that you like, compute

$$x = (1056z + 4320)/(z^3 - 15z^2 - 213z + 675)$$

and if you replace the value of $x=\beta_{\pm}$ by what you computed, the resulting cubic in $E$ has a factor $E - xz$.

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  • $\begingroup$ Unless I'm missing something, you get a different polynomial if you replace both $\beta_\pm$ and $\beta_\mp$ by the same variable $x$. $\endgroup$
    – j.c.
    Commented Mar 17, 2015 at 4:38
  • $\begingroup$ @j.c. Oh, I missed that, there are two different betas in the polynomial. I need to redo this. Thanks. $\endgroup$
    – Felipe Voloch
    Commented Mar 17, 2015 at 17:17
  • $\begingroup$ But what does it actually mean? I would understand $\beta_+ = 1 + \eta, \beta_- = 1 - \eta$ but I don't actually know the difference between $1 \pm \eta$ and $1 \mp \eta$. $\endgroup$
    – Felipe Voloch
    Commented Mar 17, 2015 at 17:19
  • $\begingroup$ I believe that there are two different polynomials under consideration, one where you always take the upper signs so that $\beta_\pm=\beta_+$ and $\beta_\mp=\beta_-$, and similarly, another polynomial corresponding to the lower signs. $\endgroup$
    – j.c.
    Commented Mar 18, 2015 at 2:17
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    $\begingroup$ Yes there are two such polynomials. My notation was not very clear. Take all upper or all lower signs for the two cases. The above solution is essentially correct with a correction that the resulting polynomial in x is the form $a {x}^3 + b {x}^2 + c x$. This factors as x times a quadratic. Solving the quadratic results in a quartic discriminant. This converts to an elliptic curve with a couple of torsion points. There is one generator point hence an infinite number of solutions which is what I was looking for. Thank you all for the suggestions on the original algebraic manipulation. $\endgroup$
    – Lorenz H Menke
    Commented Mar 21, 2015 at 0:54

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