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Can every polynomial $P(x)$ with integer coefficients we represented in the form $$ P(x) = Q^2(x)R(x), $$ where $Q(x)$ and $R(x)$ are polynomials with integer coefficients such that $R(x)$ has no repeated (complex) roots?

If this is true it should be well-known, then a reference would be helpful.

Intuition: if $\alpha$ is any complex root of $P(x)$ of even multiplicity $2k$, then factor $(x-\alpha)^k$ goes to $Q(x)$. Conversely, if $\alpha$ has odd multiplicity $2k+1$, then $(x-\alpha)^k$ goes to $Q(x)$ while $(x-\alpha)$ goes to $R(x)$. However, will $Q(x)$ and $R(x)$ have integer coefficients?

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Yes. Recall that $\mathbb{Z}[x]$ is a UFD and recall that, if $p(x)$ is irreducible in $\mathbb{Z}[x]$, then the roots of $p(x)$ are distinct.

So write $P(x) = \prod p_i(x)^{a_i}$, a product of irreducible polynomials in $\mathbb{Z}[x]$, and put $Q(x) = \prod p_i(x)^{\lfloor a_i/2 \rfloor}$ and $R(x) = \prod_{a_i \equiv 1 \bmod 2} p_i(x)$.

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  • $\begingroup$ Thank you. But what if roots of $p_i(x)$ are distinct but roots of their product $R(x)$ are not distinct? $\endgroup$ Oct 1, 2021 at 15:34
  • $\begingroup$ If $p_i(x)$ and $p_j(x)$ are distinct (non-associate) irreducible polynomials, then $p_i(x)$ and $p_j(x)$ have no roots in common, so $R(x)$ has no repeated roots. $\endgroup$ Oct 1, 2021 at 15:36
  • $\begingroup$ Thank you. This is exactly the fact I did not know (or forgot). This is equivalent to saying that every algebraic number $\alpha$ is the root of exactly one irreducible polynomial. I know that it is the root of unique (up to multiplicative constant) polynomial of lowest degree $p(x)$ (minimal polynomial). But did not know that it cannot be the root of another irreducible polynomial $q(x)$, even if we allow that $q(x)$ has higher degree. $\endgroup$ Oct 1, 2021 at 15:57
  • $\begingroup$ Ah, of course, if $p_i(x)$ and $p_j(x)$ are both irreducible, they are coprime, and therefore there exists polynomials $r_i(x)$ and $r_j(x)$ with rational coefficients such that $p_i(x)r_i(x)+p_j(x)r_j(x)=1$, hence $p_i(x)$ and $p_j(x)$ have no common root. I will now accept your answer. $\endgroup$ Oct 1, 2021 at 16:04

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