Does there exist a polynomial $P(x)$ with integer coefficients such that $P(1+\sqrt[3]{2})=1+\sqrt[3]{2}$ and $P(1+\sqrt5)=2+3\sqrt5$?
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Tags: Polynomials, algebra, polynomial
rchokler
03.08.2018 15:35
Let $Q(x)=P(x)-x$. Then we want $Q(1+\sqrt[3]{2})=0$ and $Q(1+\sqrt{5})=1+2\sqrt{5}$ First of all, $x=1+\sqrt[3]{2}\implies x^3-3x^2+3x-3=0$. Thus $Q(x)=(x^3-3x^2+3x-3)R(x)$ $x=1+\sqrt{5}\implies x^3-3x^2+3x-3=-2+5\sqrt{5}$ $R(1+\sqrt{5})=\frac{1+2\sqrt{5}}{-2+5\sqrt{5}}=\frac{52+9\sqrt{5}}{121}$ $R(x)$ clearly does not have integer coefficients since it does not act as a $\mathbb{Z}[\sqrt{5}]\to\mathbb{Z}[\sqrt{5}]$ map. Since if every polynomial in $\mathbb{Z}[x]$ that is factorable over $\mathbb{Q}$ is also factorable over $\mathbb{Z}$ with factors differing by scalar multiplying the factors, it is clear that $Q(x)$ can't have all integer coefficients, and neither can $P(x)$.
ThE-dArK-lOrD
03.08.2018 17:15
VMO 2017 P2
HKIS200543
04.08.2018 14:38
Suppose for the sake of contradiction such $P$ does exist. Like above, let $Q(x) = P(x) - x$. Note that the minimal polynomials of $1 + \sqrt[3]{2}$ and $ 1 + \sqrt{5}$ are $(x-1)^3 - 2 = x^3 - 3x^2 + 3x - 3$ and $(x-1)^2 - 5 = x^2 - 2x - 4$, respectively.
Because $Q(1 + \sqrt[3]{2}) = 0$, we have that $Q(x) = (x^3 - 3x^2 + 3x - 2)R(x)$. Since $1 + \sqrt{5}$ is a zero of $Q(x) - 2x + 1$, we have that
\[ x^2 - 2x - 4 \text{ divides } Q(x) - 2x + 1 = R(x) (x^3 - 3x^2 + 3x - 3) - 2x + 1 .\]Doing the long division, we have that
\[ R(x)(x-1) + \frac{R(x) ( - x - 3) -2x + 1}{x^2 - 2x - 4} \]is an integer polynomial. Conseqeuntly, it maps integers to integers. Plugging in $x = 0 $ gives that $R(0)$ should be odd and plugging in $x = 4$ gives $R(4)$ is divisible by $4$ and thus $R(0)$ is even, which is a contradictiom.