SAUDI ARABIAN MATHEMATICAL COMPETITIONS

A polynomial with integer coefficients is called "nice" if the coefficients in the terms whose degree is not a multiple of $3$, are multiples of $3$, and their sum is a multiple of $6$. Clearly, the sum of nice polynomials is a nice polynomial, and the sum of cube-represented polynomials is a cube-represented polynomial. We shall prove the following lemma:

Lemma. A polynomial is cube-represented if and only if it is nice.

Consider the cube of a polynomial $P(x)$ with integer coefficients, where: $$P(x)=a_n{x}^n+\cdots +a_{1}x+a_0$$ After expanding, all terms of $P^3(x)$ are either of the form $a_i^3{x}^{3i}$, or $3a_i^2{a}_j{x}^{2i+j}$ with $i$ different from $j$, or of the form $6a_i{a}_j{a}_k{x}^{i+j+k}$ with $i,j,k$ distinct.

We have $3\mid 2i+j$ if and only if $3\mid 2j+i$, and $a_i^2{a}_j+a_j^2{a}_i=a_i{a}_j(a_i+a_j)$ which is always an even number. From here, clearly a cube-represented polynomial has to be a nice polynomial, as defined above.

Second, consider a nice polynomial $P(x)$. We will prove by induction on $n$ that $P(x)$ is cube-represented, with $n$ being the integer such that the degree of $P(x)$ does not exceed $3n$, but is larger than $3n-3$.

For the base case where $n=1$, $P(x)=a_3{x}^3+3a_2{x}^2+3a_{1}x+a_0$, where $2\mid a_1-a_2$. We can write $$P(x)=\frac{a_1+a_2}{2}(x+1{)}^3+\frac{a_1-a_2}{2}(x-1{)}^3+(a_3-a_1)x^3+(a_0-a_2)1^3$$ which is obviously a cube-represented polynomial.

Assume the result is true for $n-1$. If $P(x)=a_{3n}{x}^{3n}+3a_{3n-1}{x}^{3n-1}+3a_{3n-2}{x}^{3n-2}+Q(x)$ is nice, where the degree of $Q(x)$ is at most $3n-3$.

Then either $a_{3n-1}+a_{3n-2}$ is even and $Q(x)$ is good, or $a_{3n-1}+a_{3n-2}$ is odd, and $Q(x)-3x^{3n-4}$ is good.

In the first case $$\begin{array}{rl}& P(x)=\frac{a_{3n-1}+a_{3n-2}}{2}(x^n+x^{n-1}{)}^3+\frac{a_{3n-2}-a_{3n-1}}{2}(x^n-x^{n-1}{)}^3+\\ & (a_{3n}-a_{3n-2})x^{3n}+Q(x)-a_2(x^{n-1}{)}^3\end{array}$$

In the second case $$\begin{array}{rl}P(x)& =\frac{a_{3n-1}+a_{3n-2}-1}{2}(x^n+x^{n-1}{)}^3+\frac{a_{3n-2}-a_{3n-1}-1}{2}(x^n-x^{n-1}{)}^3\\ & +(x^n+x^{n-2}{)}^3+(a_{3n}-a_{3n-2})x^{3n}+(Q(x)-3x^{3n-4})-a_2{x}^{3n-3}\end{array}$$ so in both cases, we have $P(x)$ is a nice polynomial. Our lemma has been proved.

Back to our problem,

1. With the lemma, we can easily verify that $P(x)=3x^2$ is not a nice polynomial, so it cannot be a cube-represented polynomial.

2. We need to count the number of triples $(a,b,c)$ which belong to the set $\{1,2,\dots ,2017\}$ such that $P(x)=a{x}^2+bx+c$ is cube-represented, that is, it's nice.

$P(x)$ is nice if and only if $a,b$ are multiples of $3$ and $a+b$ is a multiple of $2$.

We have $a,b\in \{3,6,\dots ,2016\}$ and $a+b$ is even, there are $336^2$ such pairs. Value $c$ can be arbitrary, so the number of triples $(a,b,c)$ is $2017\cdot 336^2$, which is the number of quadratic cube-represented polynomials satisfying the problem.