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(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?

(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?

(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,

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

using the roots $x_i$ of,

$$x^4+ax^3+bx^2+cx+d=0\tag5$$

with $v$ also a quartic root, and rational $u,w$ satisfying,

$$u^4+au^3+bu^2+cu+d=w^4\;(?)\tag6$$

By trial and error, I found that the equation,

$$x^4+x^3+m^2x+x+1=0\tag7$$

and using $u=0$ can be used on $(4)$ such that $v=t^2$ is a root of,

$$t^4 - 16 t^3 - 2 (6 m-29) t^2 + 16 (2 m-3) t + (2 m-3)^2=0$$

Example: Let $m=2$, then,

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

with complex roots $x_{1,2} \approx -0.39\pm1.84i$, and $x_{3,4} \approx -0.109\pm0.109i$. Let $u=0$ in $(4)$ so,

$$x_1^{1/4}+x_2^{1/4}+x_3^{1/4}+x_4^{1/4} = (t^2)^{1/4} = 3.65599\dots$$

where $t\approx13.3663$ is a root of

$$t^4 - 16t^3 + 34t^2 + 16t + 1 = 0$$

I don't know if there are other $u$ that can also work on this family.

III. Quintic

Question: Is it possible to find a class of quintics that is irreducible in $\mathbb Q[x]$ 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 $v$ is the root of quintic also with rational coefficients, or is there something in Galois theory that may be an obstruction?

(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?

(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 necessarily 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,

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

using the roots $x_i$ of,

$$x^4+ax^3+bx^2+cx+d=0\tag5$$

with $v$ also a quartic root, and rational $u,w$ satisfying,

$$u^4+au^3+bu^2+cu+d=w^4\;(?)\tag6$$

By trial and error, I found that the equation,

$$x^4+x^3+m^2x+x+1=0\tag7$$

and using $u=0$ can be used on $(4)$ such that $v=t^2$ is a root of,

$$t^4 - 16 t^3 - 2 (6 m-29) t^2 + 16 (2 m-3) t + (2 m-3)^2=0$$

I don't know if there are other $u$ that can also work on this family.

III. Quintic

Question: Is it possible to find a class of quintics that is irreducible in $\mathbb Q[x]$ 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 $v$ is the root of quintic also with rational coefficients, or is there something in Galois theory that may be an obstruction?

(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,

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

using the roots $x_i$ of,

$$x^4+ax^3+bx^2+cx+d=0\tag5$$

with $v$ also a quartic root, and rational $u,w$ satisfying,

$$u^4+au^3+bu^2+cu+d=w^4\;(?)\tag6$$

By trial and error, I found that the equation,

$$x^4+x^3+m^2x+x+1=0\tag7$$

and using $u=0$ can be used on $(4)$ such that $v=t^2$ is a root of,

$$t^4 - 16 t^3 - 2 (6 m-29) t^2 + 16 (2 m-3) t + (2 m-3)^2=0$$

Example: Let $m=2$, then,

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

with complex roots $x_{1,2} \approx -0.39\pm1.84i$, and $x_{3,4} \approx -0.109\pm0.109i$. Let $u=0$ in $(4)$ so,

$$x_1^{1/4}+x_2^{1/4}+x_3^{1/4}+x_4^{1/4} = (t^2)^{1/4} = 3.65599\dots$$

where $t\approx13.3663$ is a root of

$$t^4 - 16t^3 + 34t^2 + 16t + 1 = 0$$

I don't know if there are other $u$ that can also work on this family.

III. Quintic

Question: Is it possible to find a class of quintics that is irreducible in $\mathbb Q[x]$ 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 $v$ is the root of quintic also with rational coefficients, or is there something in Galois theory that may be an obstruction?