Good Numbers

Fix $n \in \mathbb{N}$ . Let $a = 2^n + \frac{1}{2}$ and $b = \frac{1}{2}$ , which are distinct non-integral real numbers.

Then, for each $1 \le k \le n$ , we have: $a^k - b^k = \left(2^n + \frac{1}{2}\right)^k - \frac{1}{2^k} = \sum_{i = 0}^{k} \binom{k}{i} 2^{n(k - i)}\frac{1}{2^i} - \frac{1}{2^k}$ $= \sum_{i = 0}^{k-1} \binom{k}{i} 2^{n(k - i)-i}$ Now in this sum we have $i \le k - 1$ . Hence, $n(k - i)-i \ge n - k + 1 \ge 1$ .

Therefore, each exponent is a natural number. Hence each term in that sum is an integer. Hence $a^k - b^k$ is an integer for all $1 \le k \le n$ .

Hence all natural numbers are good, so the answer is $\boxed{0}$ .

The wording of the question hints that there is a finite number of NOT-GOOD numbers. Let us adopt this as an assumption.

First note that if a number is GOOD then, directly from the definition, every smaller natural number is also good.

So if any number is NOT GOOD, every greater natural number must also be NOT GOOD, and so if there is at least one NOT GOOD number there must be an infinite number of NOT GOOD numbers.

Since this is contrary to our assumption, there cannot be any NOT GOOD numbers.

So the number of NOT GOOD numbers is zero.