$x=p^{k/5}+p^{l/5}+p^{m/5}+p^{n/5}$ is a root of a quintic equation for any $p \in \mathbb{Q}$ and $k,l,m,n \in \mathbb{Z}$?

Question

I first derived a special case of $k=1,~l=2,~n=3,~m=4$:

$$x=p^{1/5}+p^{2/5}+p^{3/5}+p^{4/5}$$

$$x^5-10 p x^3-10p(p+1)x^2-5p(p^2+p+1)x-p(p^3 + p^2 + p + 1)=0$$

More general case works (even though it gives much longer formulas for coefficients):

$$x=ap^{1/5}+bp^{2/5}+cp^{3/5}+dp^{4/5},~~~~a,b,c,d,p \in \mathbb{Q} $$

But after encountering some other cases, I checked with Mathematica (MinimalPolynomial) and found that much more general case also leads to quintic polynomials:

$$x=ap^{k/5}+bp^{l/5}+cp^{m/5}+dp^{n/5} \tag{1}$$

For any $k,l,m,n \in \mathbb{Z}$ - including negative exponents.

I can probably derive the polynomials for any particular combination of exponents, and I can't hope for a general formula.

But can we prove that the numbers defined by $(1)$ are always roots of some quintic (degree $5$) polynomial with rational coefficients?

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As a random example, we have:

$$x=13^{-11/5} - 3 \cdot 13^{-13/5} + 2 \cdot 13^{3/5} - 7 \cdot 13^{19/5}=-119687.54649388942204$$

Which is a root of:

$$302875106592253 x^5+53764811821110 x^3+98398475724403213219380 x^2+ \\ +24203484201970417908238969616010 x +7441789985038797830837836893979741390939$$

Answer 1

We prove the following slightly more general statement:

For $a,b,c,d,e\in\mathbb Q$, if $x=ap^{1/5}+bp^{2/5}+cp^{3/5}+dp^{4/5}+e$, then $x$ has a minimal polynomial of degree at most $5$.

(By moving powers of $p$ into/out of the coefficient, we can restrict the exponents $k,l,m,n$ to the set $\{0,1,2,3,4\}$.)

Consider the vector space $V$ with basis $\{1,p^{1/5},p^{2/5},p^{3/5},p^{4/5}\}$ over $\mathbb Q$. Then each element of $S = \{1,x,x^2,x^3,x^4,x^5\}$ can be interpreted as a vector in $V$. Since $V$ is a vector space of dimension $5$, set $S$ cannot be linearly independent--thus some linear combination of these vectors must equal $0$. This gives us a corresponding polynomial of degree at most $5$.

Answer 2

It becomes easier to understand if we embed the original expression in an appropriate group of five:

1) $ap^{1/5}+bp^{2/5}+cp^{3/5}+dp^{4/5}$

2)$a\omega p^{1/5}+b\omega^2p^{2/5}+c\omega^3p^{3/5}+d\omega^4p^{4/5}$

3)$a\omega^2p^{1/5}+b\omega^4p^{2/5}+c\omega p^{3/5}+d\omega^3p^{4/5}$

4)$a\omega^3p^{1/5}+b\omega p^{2/5}+c\omega^4p^{3/5}+d\omega^2p^{4/5}$

5)$a\omega^4p^{1/5}+b\omega^3p^{2/5}+c\omega^2p^{3/5}+d\omega p^{4/5}$

Here $\omega$ is one of the non-real fifth roots of unity (all four choices work equally well). We take the usual sum-product cominatoons of these roots to get their polynomial, and all the terms containing $\omega$ are killed by the relationships

$1+\omega+\omega^2+\omega^3+\omega^4=0$

$\omega^5=1$

We are then left with only rational coefficients, and there we are.

Answer 3

Characteristic polinomial of matrix: $$\begin{pmatrix}a & gm & dm & cm & bm\\ b & a & gm & dm & cm\\ c & b & a & gm & dm\\ d & c & b & a & gm\\ g & d & c & b & a\end{pmatrix}$$

has a root: $$x=g\,{{m}^{4/5}}+d\,{{m}^{3/5}}+c\,{{m}^{2/5}}+b\,{{m}^{1/5}}+a$$