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Given a positive integer $m>1$, prove that there exists a polynomial of integer coefficients $P(x)$ that is irreducible on $\mathbb{Z}[x]$, has degree $2018$ and has a real root $r$ satisfying $m \mid \big(1+ \lfloor r^n \rfloor \big)$, for all positive integer $n$.
$ \big \lfloor x \rfloor$ is the floor function of $x$.

This question was posted on Aops last year without any progress. I also contacted the OP to know the source but didn't get any answer.

Is this result true? Is it possible to construct this polynomial explicitly?
Can this result be generalized to other degrees?

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    $\begingroup$ It's likely an exercise on the standard trick of an algebraic number, all of whose conjugates have absolute value less than one (Pisot–Vijayaraghavan number), if in addition all the conjugates are positive, then the floor of $1+r^n$ is equal to the sum of $n$'th powers, which can also be made divisible by $m$. Not sure if this is appropriate for MO, the trick is standard but there is a lot of details to fill in. $\endgroup$
    – Aleksei Kulikov
    Commented Aug 31 at 13:59
  • $\begingroup$ @AlekseiKulikov Could you please give some more details? How do you make the sum of $n$'th powers divisible by $m$? If all conjugates are positive how is it possible to ensure that $P(x)$ has only integer coefficients? $\endgroup$
    – jack
    Commented Aug 31 at 19:08

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I will prove it modulo a fact written in Wikipedia which I don't know how to prove (but since, as we all know, Wikipedia is the universal arbiter of truth and is never wrong, I will equate it with complete proof in my heart).

Recall that an algebraic integer $u=u_1$ is called a Pisot–Vijayaraghavan number if $|u| > 1$ and all its conjugate $u_2, u_3, \ldots, u_n$ satisfy $|u_k| < 1$. On the Wikipedia page about these numbers, it is written that there exists such a number in any real algebraic number field, and it generates said field, so we will pick our favourite real algebraic number field (together with its embedding into $\mathbb{R}$) of degree $n = 2018$ and take said $u$. It will be a root of a polynomial with integer coefficients and leading coefficient one.

First, we will replace $u$ with $s = s_1 = u^2$. It is still a Pisot–Vijayaraghavan number (its conjugates are squares of conjugates of $u$), and it is also almost immediate that its degree is still $n$ (otherwise, if its degree is $n/2$, then both $u$ and $-u$ must be conjugate and lie in our field, but then $|-u| > 1$ -- contradiction to $u$ being a Pisot–Vijayaraghavan number). Now, $s$ and all its conjugates $s_2, s_3, \ldots, s_n$ are positive.

The powers of $s$ are also Pisot–Vijayaraghavan numbers, and there are arbitrarily high powers of $s$ which also have degree $n$ (this second fact is true for any algebraic integer). Let us pick $k$ big enough such that $s^k$ has degree $n$ and $(\max_{l = 2, \ldots , n} |s_l|)^k < \frac{1}{nm}$. Note that since all $s_l$ are between $0$ and $1$, such $k$ exists. Finally, put $r =r_1= ms^k$, we will show that it works. We begin with the proposition that $$1 + \lfloor r^p\rfloor = r_1^p + r_2^p + \ldots + r_n^p.$$

Indeed, the right-hand side is an integer, since $r_l$ are conjugate to each other and are algebraic integers, it is strictly bigger than $r_1^p$ (here we used that $n>1$!), and also it is smaller than $r_1^p + 1$ -- this is equivalent to the fact that $r_2^p + \ldots + r_n^p < 1$, which is true since $$r_2^p + \ldots + r_n^p \le r_2 + \ldots + r_n \le \frac{m(n-1)}{nm} =\frac{n-1}{n} < 1.$$

Finally, about divisibility. We have $$r_1^p + \ldots + r_n^p = m^p(s_1^{kp}+\ldots + s_n^{kp}),$$ and the sum in the brackets is also an integer since $s_l$ are conjugate algebraic integers. So, $1 + \lfloor r^p\rfloor$ is even divisible by $m^p$, hence by $m$.

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    $\begingroup$ Careful: the conjugates $s_2,\ldots,s_n$ won't automatically be real unless you start with a totally real algebraic number field. If I got my bearings right, there's (for example) a subfield of the field of 12109th roots of unity that should work. $\endgroup$
    – GNiklasch
    Commented Sep 1 at 13:15
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    $\begingroup$ Addendum: Here's how to prove the "elementary fact" from Wikipedia. Let $\sigma_i: K\to \mathbb{C}$ be the embeddings of $K$, with $\sigma_1$ real. Then complex conjugation acts on $\sigma_2,\ldots,\sigma_n$, and we get a joint embedding $\mathcal{O}_K\to (\prod_{i\neq 1} \mathbb{C})^{C_2}$. The target is a real vector space of dimension $n-1$, while $\mathcal{O}_K$ is a lattice of rank $n$. So its image cannot be discrete, and we find nonzero $\alpha \in \mathcal{O}_K$ whose image under the restricted embedding has arbitrarily small norm. (cont. in next comment) $\endgroup$
    – Achim Krause
    Commented Sep 2 at 8:50
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    $\begingroup$ In particular, we find $\alpha$ with $|\sigma_i(\alpha)|<1$ for all $i\neq 1$, and so $|\sigma_1(\alpha)|>1$ since their product is a nonzero integer. So it is a PV number. Finally, it generates $K$. Indeed, $[K:\mathbb{Q}(\alpha)]$ is $1$, otherwise the embedding $\sigma_1|_{\mathbb{Q}(\alpha)}$ would extend to $K$ in multiple ways, so there would be more than one embedding with $|\sigma_i(\alpha)|>1$. $\endgroup$
    – Achim Krause
    Commented Sep 2 at 8:52

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