Basic Mathematics & Logarithm Question - None

Question

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 $\lim_{N \to \infty} \dfrac{S(N)}{N^\alpha} = \beta$.

Accepted Answer

Step 1: Characterize squarish integers. An integer n is squarish if n = k² (perfect square) or |n - k²| = m² for some perfect squares k² and m².

Step 2: Count squarish integers between consecutive perfect squares. Between k² and (k+1)², squarish integers are those within distance m² of k² or (k+1)². For large k, approximately 2k integers lie within distance k of k², giving roughly 2k squarish integers per interval [k², (k+1)²].

Step 3: Sum over all perfect squares up to N. With √N perfect squares up to N, the total count is roughly:
S(N) ~ ∑_{k=1}^{√N} 2k ~ ∫₁^{√N} 2k dk = [k²]₁^{√N} ~ N

Step 4: Refine the asymptotic analysis. More careful analysis accounting for the length of intervals [k², (k+1)²] ≈ 2k shows:
S(N) ~ ∫₁^{√N} 2√k dk = [\frac{4k^{3/2}}{3}]₁^{√N} ~ \frac{4}{3}N^{3/4}

Step 5: Extract α and β. Comparing S(N) ~ βN^α:
∴ α = \boxed{\frac{3}{4}}, β = \boxed{\frac{4}{3}}

Question:

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