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Is there a polynomial $P(x)$ with integral coefficients s.t. $P(1+\sqrt[3]{2})=1+\sqrt[3]{2}$ and $P(1+\sqrt{5})=2+3\sqrt{5}$. I'm not good at this kind of problem, all I did up to now is just assuming $deg P=3,4,5$ and if I'm not wrong, $degP \geq6$. Help me

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    Funny my friend @GiangNam said he can solve this in 8 lines – 2017-01-06
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    Perhaps your friend can share his solution here. – 2017-01-06
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    You have accepted an answer that does not seem to have answered your question. – 2017-01-06

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Let $f(x)=(x-1)^3-2$, $g(x)=(x-1)^2-5$. The numbers $a=1+\sqrt[3]2$ and $b=1+\sqrt 5$ are roots respectively of $f$ asnd $g$.

It is a lot of work, but now you can solve the following system of congruences, using the Chinese remainder theorem applied to the Euclidean domain $\Bbb Q[x]$:

$$\begin{cases}P(x)\equiv x&\pmod f\\P(x)\equiv3x-1&\pmod g\end{cases}$$

So you get $$P(a)=h_1(a)f(a)+a=a$$ $$P(b)=h_2(b)g(b)+3b-1=3b-1$$ as desired.

As it has been suggested in comments, $P$ may have non integer coefficients. In fact, using software I have found $$P(x)=\frac1{121}(9x^4+16x^3-102x^2+223x-129)$$

I don't know if it is possible to find a polynomial $h_3(x)$ such that $$P(x)+h_3(x)f(x)g(x)\in\Bbb Z[x]$$

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    Very smart and elegant solution +1 – 2017-01-06
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    Why does $P$ have integer coefficients? – 2017-01-06
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    Why do we have to use Q[x]. and are maximal ideal in Z[x],right? – 2017-01-06
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    @ajotatxe why not just Z[x] – 2017-01-06
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    @TrầnNguyênTry Because $\Bbb Z[x]$ is not an Euclidean domain. – 2017-01-06
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    @ajotatxe we need that? Chinese remaider theorem only needs comaximal ideals, right? – 2017-01-06
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    @TrầnNguyênTry, see http://math.stackexchange.com/questions/1591430/denominator-in-rational-gcd-of-integer-polynomials for instance. – 2017-01-06