There could be some errors as computations ain't a joke. Let $Q:=P(X)-X \in \mathbb{Z}[X]$ than $Q(1+\sqrt[3]{2})=0$ and $Q(1+\sqrt{5})=1+2\sqrt{5}$.Now notice the polynomial $R(x)=x^3-3x^2+3x-3=(x-1-\sqrt[3]{2})(x-1-\sqrt[3]{2} \omega_1)(x-1-\sqrt[3]{2} \omega_2)$,it irreducibile over $Z[X]$ and $(R,Q)\not =1$ $\implies$ $R\mid Q$.Let $Q=RS$ where of course $S\in \mathbb{Z}[X]$.We have: $$R(1+\sqrt{5})S(1+\sqrt{5})=1+2\sqrt{5}$$Taking norms of left side we have $-121\cdot N(S(1+\sqrt{5}))$ and $N(1+2\sqrt{5})=-19$ but as $S\in \mathbb{Z}[x]$ we have $121\mid 19$,absurd so no such polynomial.

Pretty similar idea is used here

Similar to Zhautykov Olympiad

nikolapavlovic wrote: There could be some errors as computations ain't a joke. Let $Q:=P(X)-X \in \mathbb{Z}[X]$ than $Q(1+\sqrt[3]{2})=0$ and $Q(1+\sqrt{5})=1+2\sqrt{5}$.Now notice the polynomial $R(x)=x^3-3x^2+3x-3=(x-1-\sqrt[3]{2})(x-1-\sqrt[3]{2} \omega_1)(x-1-\sqrt[3]{2} \omega_2)$,it irreducibile over $Z[X]$ and $(R,Q)\not =1$ $\implies$ $R\mid Q$.Let $Q=RS$ where of course $S\in \mathbb{Z}[X]$.We have: $$R(1+\sqrt{5})S(1+\sqrt{5})=1+2\sqrt{5}$$Taking norms of left side we have $-121\cdot N(S(1+\sqrt{5}))$ and $N(1+2\sqrt{5})=-19$ but as $S\in \mathbb{Z}[x]$ we have $121\mid 19$,absurd so no such polynomial. I don't really understand why $R \mid Q$?? Can you explain it in a more specific way?

$\mathbb{Z}[X]$ is an ufd and hence as $R$ is irreducible it's also a prime element and hence as (isomorphic to the NT in $\mathbb{Z}$) $(R,S)\not =1$ we have $R\mid S$.

In the same vein, we are solving a system of linear congruences over quotient rings of $\mathbb Q[x]$ to see if the solution has integral coefficients (which it doesn't). Edit: I'm an idiot! The above actually leads to a very nice and computationally friendly approach: Instead of working in $\mathbb Q[x]$, we work over $\mathbb Z[x]$ and apply some basic results of commutative algebra. Specifically, let $\mathfrak{a}=(x^2-2x-4)$ and $\mathfrak{b}=(x^3-3x^2+3x-3)$. By the generalized Chinese Remainder Theorem, the congruence has not solutions (which, if only for reasons of avoiding computation, we hope is the case) if and only if 1: $\mathfrak{a}+\mathfrak{b}\subsetneq \mathbb Z[x]$, and 2: $3x-1\not \equiv x\pmod{\mathfrak{a}+\mathfrak{b}}\iff 2x-1\not\in \mathfrak{a}+\mathfrak{b}$. To that end, consider $\mathfrak{I}=\mathbb Z\cap(\mathfrak{a}+\mathfrak{b})$, which can be seen to be an ideal of $\mathbb Z$, and thus is generated by a single principle element. Performing the "GCD" algorithm, but with all coefficients in $\mathbb Z$ as opposed to $\mathbb Q$, strongly suggests that $\mathfrak{I}=(121)$; this is a good sign, as it supports part 1/2 of our conjecture. We proceed to a slightly ad hoc step: consider the (proper!) ideals of $\mathbb Z[x]$ of the form $\mathfrak{J}_{a,n}=\{P\in\mathbb Z[x]:P(a)\in (n)\}$ where $a,n\in\mathbb Z$ and $n$ is not a unit. In particular, we conjecture that there is some integer $a$ for which $\mathfrak{a},\mathfrak{b} \subseteq \mathfrak{I}_{a,11}$, and $2x-1\not \in \mathfrak{J}_{a,11}$$^{[1]}$. We can easily compute $x^2-2x-4$ to vanish over $\mathbb Z/(11)$ at $5$ and $-3$; cross-checking with $x^3-3x^2-3x+3$ shows that $a=-3$ works beautifully. We are done $\blacksquare$ [1]: While it's more natural to pick $n=121$, any satisfactory pair $(a,121)$ works just as well as $(a,11)$, and it's loads easier from a computational perspective to work over $\mathbb Z/(11)$ than it is to work over $\mathbb Z/(121)$

Here is my solution for this problem Solution Let $Q(x) = P(x + 1) - 1$ then $Q(x)$ has integer coefficients We have: $Q(\sqrt[3]{2}) = \sqrt[3]{2}$, $Q(\sqrt{5}) = 3\sqrt{5}$ But: $x^3 - 2$ is a polynomial which has integer coefficients and smallest degree so: $Q(x) - x \vdots x^3 - 2$ Let $Q(x) - x = (x^3 - 2)R(x)$ then: $R(x)$ has integer coefficient So: $R(\sqrt{5}) = a + b\sqrt{5}$ with $a; b \in \mathbb{Z}$ Then: $(a + b\sqrt{5})(5\sqrt{5} - 2) = 2\sqrt{5}$ or $(5a - 2b - 2)\sqrt{5} + 25b - 2a = 0$ So: $5a - 2b = 2$ and $25b - 2a = 0$ or $a = \dfrac{50}{121}$, $b = \dfrac{4}{121}$, conflict with $a; b \in \mathbb{Z}$ Hence: there doesn't exist $P(x)$ which has integer coefficient satisfies the problem