Problem 1537

1. Define the Polynomial:
Consider the polynomial $P(x) = \sum_{n \in N} 2x^n$, where $N$ is a set of 1007 positive integers. This polynomial is constructed such that each term represents a digit in the number $n$.

2. Ensure Unique Sums:
We need to ensure that all sums of 1, 2, or 3 elements in $N$ are unique. This can be achieved by choosing $N$ to be the first 1007 powers of 9001. This choice ensures that the sums of any 1, 2, or 3 elements are distinct because the powers of 9001 grow exponentially.

3. Square of the Polynomial:
When we consider $P(x)^2$, the terms are either $4x^{2n}$ or $8x^{n_1+n_2}$. This follows from the expansion using the multinomial theorem. Specifically:
$$P(x)^2 = \left( \sum_{n \in N} 2x^n \right)^2 = \sum_{n \in N} 4x^{2n} + \sum_{n_1, n_2 \in N, n_1 \neq n_2} 8x^{n_1+n_2}$$
Since the sums $n_1 + n_2$ are unique, the resulting polynomial will have terms with coefficients 4 or 8, which are sweet digits.

4. Cube of the Polynomial:
When we consider $P(x)^3$, the terms are either $8x^{3n}$, $24x^{2n_1+n_2}$, or $48x^{n_1+n_2+n_3}$. This follows from the expansion using the multinomial theorem. Specifically:
$$P(x)^3 = \left( \sum_{n \in N} 2x^n \right)^3 = \sum_{n \in N} 8x^{3n} + \sum_{n_1, n_2 \in N, n_1 \neq n_2} 24x^{2n_1+n_2} + \sum_{n_1, n_2, n_3 \in N, n_1 \neq n_2 \neq n_3} 48x^{n_1+n_2+n_3}$$
Again, since the sums $2n_1 + n_2$ and $n_1 + n_2 + n_3$ are unique, the resulting polynomial will have terms with coefficients 8, 24, or 48, which are sweet digits.

5. **Choosing $x$ and Ensuring Sweetness:**
By setting $x = 100$, we avoid overlapping of two-digit products. This ensures that the polynomial $P(100)$ represents a number whose digits are sweet. The sum of the digits of $n$ is 2014, as required.

6. Conclusion:
Therefore, the number $P(100)$ is a sweet number whose square and cube are also sweet.

$\blacksquare$