SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, we shall prove two following lemmas:

Lemma 1. If $a$ is an integer number that is not a perfect cube, and $m,n,p$ are integer numbers such that $m+n\sqrt[3]{a}+p\sqrt[3]{a^2}=0$, then $$m=n=p=0.$$ Proof of Lemma 1. Since $m+n\sqrt[3]{a}+p\sqrt[3]{a^2}=0$, then $\sqrt[3]{a}$ is the root of $P(x)=p{x}^2+nx+m$ with integer coefficients. Consider the polynomial $Q(x)=x^3-a$, then $\sqrt[3]{a}$ is also a root of $Q(x)$. Assume that $Q(x)$ is reducible, then one of the factors must have degree 1, that is $Q(x)$ has a rational root. But we know that $\sqrt[3]{a}$ is irrational since $a$ is not a perfect cube (by assumption). Hence, $Q(x)$ is irreducible. In other words, $Q(x)$ is the minimal polynomial of $\sqrt[3]{a}$, and this leads to $P(x)=0$, or $m=n=p=0$.

Lemma 2. Given that $P(x)$ is a polynomial with integer coefficients, and $a$ is an integer such that $\sqrt[3]{a}$ is not an integer. If $$P\left(\sqrt[3]{a^2}+\sqrt[3]{a}\right)=m+n\sqrt[3]{a}+p\sqrt[3]{a^2}$$ then $a-1\mid n-p$.

Proof of Lemma 2. We will prove by induction on the degree of $P(x)$ which is denoted by $k$. For $k=1$, it is trivial. Assume that the statement is true for $k=t$, consider $P(x)$ of degree $t+1$, we write $P(x)=xQ(x)+s$, with $Q(x)$ of degree $t$ and $s=P(0)$ is an integer. Assume that $Q\left(\sqrt[3]{a^2}+\sqrt[3]{a}\right)=d+e\sqrt[3]{a}+f\sqrt[3]{a^2}$, then by induction hypothesis $a-1\mid e-f$. This gives $$\begin{array}{rl}P\left(\sqrt[3]{a^2}+\sqrt[3]{a}\right)& =\left(\sqrt[3]{a^2}+\sqrt[3]{a}\right)\left(d+e\sqrt[3]{a}+f\sqrt[3]{a^2}\right)+s\\ & =s+ea+fa+(d+fa)\sqrt[3]{a}+(d+e)\sqrt[3]{a^2}\end{array}$$ Hence, $(d+fa)-(d+e)=fa-e=(fa-f)+(f-e)$, which is a multiple of $a-1$.

Now we turn to the problem.

For part a), we show that $P(x)$ does exist, for example we choose $P(x)=u{x}^2+vx+w$, where $u,v,w$ must satisfy $$u+v=220,5u+v=284,10u+w=0.$$ That is, $P(x)=16x^2+204x-160$.

For part b), since $1210-1184=26$, which is not a multiple of $4$, in this case there is no $P(x)$ satisfying the condition of the problem.