The sum of the $n^{\text{th}}$ powers of the roots of a monic polynomial with integer coefficents is always integer

inductionpolynomialssolution-verification

The following question is inspired by

Proving that the sum of $n^{th}$ powers of the roots of a quadratic with integer coefficients is also an integer

I am curious, if we can generalize it to all monic polynomials with integer coefficents, i.e. is the sum of the $n^{\text{th}}$ powers of the roots of a monic polynomial with integer coefficents always integer?

Here is what i have done so far:

Let $p(x)=x^d+\sum_{i=0}^{d-1}c_i\cdot x^i$ with $c_i\in \mathbb Z$. Let $\alpha$ be an arbitrary root of $p$, then from $\alpha^n+\sum_{i=0}^{d-1}c_i\cdot\alpha^i=0$ follows
$$
\alpha^{d}=-\sum_{i=0}^{d-1}c_i\cdot\alpha^i
$$

Hence, for $n\geq d$, we can multiply by $\alpha^{n-d}$ and get
$$
\alpha^{n}=-\sum_{i=0}^{d-1}c_i\cdot\alpha^{n-d+i}.
$$

Since this holds for all roots, we can sum up this equation over all roots of $p$ but with a fixed $n$. Let $s_n$ be the sum of the $n^{\text{th}}$ power of all roots of $p$, we get for $n\geq d$
$$
s_n = -\sum_{i=0}^{d-1}c_i\cdot s_{n-d+i},
$$

which is an integer, if $s_0, \ldots, s_{d-1}$ are integers. And here I stuck, because I can't prove, that they are integers. I know, that $s_0=d$ and $s_1=(-1)^{d+1}c_1$ are integers, but nothing more.

Best Answer

The answer is yes.

Let $Q=X_1^n + ... + X_d^n \in \mathbb{Z}[X_1,...,X_d]$. The polynomial $Q$ is a symmetric polynomial, so by the fundamental theorem of symmetric polynomials, $Q$ is a polynomial in the elementary symmetric polynomials $e_1, ..., e_d$. So if $\lambda_1, ..., \lambda_d$ denote the roots of $p$, you get that $Q(\lambda_1, ..., \lambda_d)$ is a polynomial (with integer coefficients) in the $e_k(\lambda_1, ..., \lambda_d)$, which are exactly the coefficients of $p$. So $Q(\lambda_1, ..., \lambda_d)$ is a polynomial in the coefficients of $p$, and because $p$ has integer coefficients, then $Q(\lambda_1, ..., \lambda_d) \in \mathbb{Z}$.

Notice that this works for any symmetric polynomial instead of $Q$.

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