USAMTS 5/2/36

Proof: Assume that there is a polynomial $P(x)$ of degree $n$, with integer coefficients, and without trivial root of $0$, such that
$$P(\sqrt[3]{5}+\sqrt[3]{25})=2\sqrt[3]{5}+3\sqrt[3]{25}$$
Therefore, $P(x)$ can be expressed as $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots a_2x^2+a_1x+a_0$$
where $n\ge 0$, $a_i$ are integers, $i=0,1,2,\ldots, n$, and $a_0\ne 0$, and $a_n\ne 0$.

Let $t=\sqrt[3]{5}+\sqrt[3]{25}$, then $$P(t)=2\sqrt[3]{5}+3\sqrt[3]{25}$$ Because $$t^3=(\sqrt[3]{5}+\sqrt[3]{25})^3=5+15(\sqrt[3]{5}+\sqrt[3]{25})+25=30+15t$$
Therefore the minimum polynomial equation with a root as $t=\sqrt[3]{5}+\sqrt[3]{25}$ is $$x^3-15x-30=0$$ By Euclidean division of polynomials with integer coefficients, for divisor $x^3-15x-30$, there exists a unqiue $Q(x)$ and a unique $R(x)$ so that
$$P(x)=(x^3-15x-30)Q(x)+R(x)$$
where $Q(x)$ is a polynomial of degree $n-3$ with integer coefficients, and $R(x)$ is a
polynomial of degree at most $2$ with integer coefficients. Note that if $n<3, Q(x)=0$.

Because $t=\sqrt[3]{5}+\sqrt[3]{25}$ is a root of polynomial equation $x^3-15x-30=0$, we have $$P(t)=(t^2-15t-30)Q(t)+R(t)=0\cdot Q(t)+R(t)=R(t)=2\sqrt[3]{5}+3\sqrt[3]{25}$$ Let $$R(x)=ax^2+bx+c$$ where $a$, $b$, and $c$ are integers, which also covers the cases when degree $n$ is $0$, $1$, or $2$, we have

$$\begin{array}{ccccl}
P(t)&=&R(t)&=&a(\sqrt[3]{5}+\sqrt[3]{25})^2+b(\sqrt[3]{5}+\sqrt[3]{25})+c \\
& & &=&a(\sqrt[3]{25}+10+5\sqrt[3]{5})+b(\sqrt[3]{5}+\sqrt[3]{25})+c \\
& & &=&(5a+b)\sqrt[3]{5}+(a+b)\sqrt[3]{25}+(10a+c) \\
& & &=&2\sqrt[3]{5}+3\sqrt[3]{25}
\end{array}$$

Because $\sqrt[3]{5}$ and $\sqrt[3]{25}$ are irrational numbers that do not share any rational common factors, in order to satisfy the following equation:

$$(5a+b)\sqrt[3]{5}+(a+b)\sqrt[3]{25}+(10a+c)=2\sqrt[3]{5}+3\sqrt[3]{25}$$ where $a+b$, $5a+b$, $10a+c$ are integers, which are rationals, we must have the following:

$$
\left\{
\begin{array}{rcrcrc@{\qquad}l}
5a & + & b & & & = & 2 \\
a & + & b & & & = & 3 \\
10a & & & + & c & = & 0
\end{array}
\right.
$$

Solving the above equations, we have $a=-\dfrac{1}{4}$, $b=\dfrac{13}{4}$, and $c=\dfrac{5}{2}$. Hence,

$$R(x)=-\dfrac{1}{4}x^2+\dfrac{13}{4}x+\dfrac{5}{2}$$ However, the above result contradicts the assumption that $R(x)$ is a polynomial with integer coefficients, nor does it work if degree $n$ is $0$ or $1$. Additionally, because $$P(x)=(x^3-15x-30)Q(x)+R(x)$$ where $Q(x)$ is a polynomial of degree $n-3$ with integer coefficients, it also contradicts the assumption that $P(x)$ is a polynomial with integer coefficients.

Therefore, the original assumption is false, which completes the proof that there is no polynomial $P(x)$ with integer coefficients such that
$$P(\sqrt[3]{5}+\sqrt[3]{25})=2\sqrt[3]{5}+3\sqrt[3]{25}$$$\boxed{\phantom{\ }}$