14
$\begingroup$

(Major revision to incorporate new results in this MO cubic version.)

Note: All coefficients are in the rationals.

I. Cubic

In the linked post, it was shown that given a general cubic (via its depressed form),

$$x^3+px+q=0\tag1$$

and its roots $x_i$, then it is possible to find infinitely many rational $u$,

$$G_3 := (u \pm x_1)^{1/3}+ (u \pm x_2)^{1/3}+ (u \pm x_3)^{1/3} = {v}^{1/3}\tag2$$

such that $v$ is also a cubic root or even rational. Apparently, a necessary but not sufficient condition is that $u$ should be a rational point on,

$$u^3+pu+q=w^3\tag3$$

which, after a transformation, is an elliptic curve.

II. Quartic

By analogy, the quartic version should be,

$$G_4:=(u +x_1)^{1/4}+ (u +x_2)^{1/4}+ (u +x_3)^{1/4}+(u+x_4)^{1/4} = {v}^{1/4}\tag4$$

where the $x_i$ are the four solutions of a quartic and $v$ is at most a quartic root. By trial and error, I found the equation,

$$x^4+x^3+\big(\tfrac{n-3}{2}\big)^2x^2+x+1=0\tag5$$

and using $u=0$, then the $RHS$ of $(4)$ is $v=t^2$ which is a root of,

$$(t^2 - 8t + 6 - n)^2 = 4 n t^2\tag6$$

Thus $v$ is a quartic but, by judicious choice of $n$, can be a quadratic or even just a linear root.

Example:

Let $n=81$, then we have the irreducible,

$$x^4 + x^3 + 1521x^2 + x + 1=0$$

with complex roots $x_{1,2} \approx -0.49\pm38.99i$, and $x_{3,4} \approx -0.0003\pm0.025i$. Let $u=0$ in $(4)$ then,

$$x_1^{1/4}+x_2^{1/4}+x_3^{1/4}+x_4^{1/4} = (225)^{1/4}$$

or,

$$x_1^{1/4}+x_2^{1/4}+x_3^{1/4}+x_4^{1/4} = (413+52\sqrt{61})^{1/4}$$

depending on the $4$th roots used.

III. Quintic

Question: Is it possible to find a class of quintics that is irreducible (over $\mathbb{Q}$) yet have roots $x_i$ such that,

$$G_5:=x_1^{1/5}+x_2^{1/5}+\dots+x_5^{1/5} = v^{1/5}$$

and $v$ at most is a root of a quintic also with rational coefficients, or is there something in Galois theory that may be an obstruction?

$\endgroup$
  • $\begingroup$ Your use of "degree of an algebraic number" is inconsistent. I guess you mean that number generate an extension of that degree, so you should say $F_3$ is a priori of degree at most or dividing 9, and similarly for $F_5$. Also, maybe it would be helpful to say what you tried. Did you try a computer search? Roots of quintics seem complicated to me, and this is what I would suggest. $\endgroup$ – Kimball Jan 15 '15 at 0:38
  • 9
    $\begingroup$ So you're looking for something like the roots of $x^5 + 6x^4 - x^3 - 32x^2 + 16x - 1$ (which are in ${\bf Q}(\cos \pi/11)$ and have fifth roots that sum to $v^{1/5}$ with $v^5 - 1370v^4 + 518385v^3 + 319010v^2 + 143005v + 1 = 0$)? $\endgroup$ – Noam D. Elkies Aug 1 '16 at 4:43
  • $\begingroup$ @NoamD.Elkies: Yes, great result! I've elaborated on your comment and gave more details as an answer below. I hope you don't mind. $\endgroup$ – Tito Piezas III Aug 2 '16 at 19:34
  • $\begingroup$ I thought you wanted me to post an answer including an explanation of why these identities exist. And yes, as I wrote in e-mail there are similar things for 7th roots and beyond, in fields with either a cyclic Galois group (such as the one contained in ${\bf Q}(\cos 2\pi/29)$) or dihedral Galois (I gave the examples of $x^5 + x^4 - 3x^3 - 3x^2 + 4x - 1$ and $x^7 - 26x^6 + 350x^5 - 2192x^4 + 6287x^3 - 6718x^2 - 162x - 1$ in the same email). $\endgroup$ – Noam D. Elkies Aug 2 '16 at 20:03
  • $\begingroup$ @NoamD.Elkies: Oops, I just saw your email. I've deleted my answer, and I'll wait for the one you planned that explains the general case. My sincere apologies. $\endgroup$ – Tito Piezas III Aug 2 '16 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.