I think my solution must be wrong because it's way too simple and it disagrees with other solutions I have seen online. But I really can't figure out where the mistake is.
Define a positive integer $n$ to be squarish if either $n$ is itself a perfect square or the distance from $n$ to the nearest perfect square is a perfect square. For example, $2016$ is squarish, because the nearest perfect square to $2016$ is $45^2=2025$ and $2025-2016=9$ is a perfect square. (Of the positive integers between $1$ and $10,$ only $6$ and $7$ are not squarish.)
For a positive integer $N,$ let $S(N)$ be the number of squarish integers between $1$ and $N,$ inclusive. Find positive constants $\alpha$ and $\beta$ such that
$$\lim_{N\to\infty}\frac{S(N)}{N^{\alpha}}=\beta,$$
or show that no such constants exist.
Since $S(N)$ just counts some of the numbers between $1$ and $N$, we have $ 0 \le S(N) \le N$ for any $N$. So for any $N$,
$$\dfrac {0}{N^{\alpha}} \le \frac{S(N)}{N^{\alpha}} \le \dfrac {N}{N^{\alpha}}$$
$$0 \le \frac{S(N)}{N^{\alpha}} \le \dfrac {1}{N^{\alpha-1}}$$
Let $\alpha = 2$
$$0 \le \frac{S(N)}{N^{2}} \le \dfrac {1}{N}$$
By the Squeeze (Sandwich?) Theorem,
$$\lim_{N\to\infty}\frac{S(N)}{N^{2}}=0$$
So $\alpha = 2$, $\beta = 0$ (or $\alpha=$ any number $>2, \beta = 0$)