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 – Jan 06 '17 at 12:12
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Perhaps your friend can share his solution here. – lhf Jan 06 '17 at 15:17
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You have accepted an answer that does not seem to have answered your question. – lhf Jan 06 '17 at 15:43
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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 $$P(x)+h_3(x)f(x)g(x)\in\Bbb Z[x]$$
ajotatxe
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@ajotatxe we need that? Chinese remaider theorem only needs comaximal ideals, right? – Jan 06 '17 at 15:40
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@TrầnNguyênTry, see http://math.stackexchange.com/questions/1591430/denominator-in-rational-gcd-of-integer-polynomials for instance. – lhf Jan 06 '17 at 15:43