Rewriting a sum of harmonic powers as a minimal polynomial

Question

Revisiting one of my older questions, I've decided to try to tackle a simpler version of the problem, this time without the square root coefficients.

Let $x_0$ be a real number such that it satisfies the equation $$x_0+x_0^{1/2}+x_0^{1/3}+\cdots+x_0^{1/n}=1$$ for a natural number $n$. What is the minimal polynomial in $\mathbb{Z}[x]$?

Of course, this is possible by brute force: isolating the smallest power of $x$ then raising both sides by its reciprocal and repeating, but it becomes extremely tedious to do when $n$ is large. Also, this does not guarantee that the polynomial obtained is minimal.

This works fine for $n=1,2,3$. The minimal polynomials are, respectively, $$x-1,\quad x^2-3x+1,\quad x^5-8x^4+24x^3-21x^2+10x-1$$ and it may be interesting to note that the sign of the coefficients are alternating.

Is there an efficient way of doing this for the general case?

The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern. — Torsten Schoeneberg, Aug 28 '18 at 10:50
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Aug 29 '18 at 18:10

score 1 · Answer 1 · answered Aug 28 '18 at 07:34

Let $x=y^L$, where $L$ is the least common multiple of $2,3,\dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^{L/2}+y^{L/3}+\cdots+y^{L/n}-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.

The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2L\times2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.

It looks like the first equation is trivial and the second equation is just raising each term to a power of $\text{lcm}$. However, this still avoids the method of finding a minimal polynomial. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Sep 01 '18 at 13:21

Rewriting a sum of harmonic powers as a minimal polynomial

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