Squarish numbers

Number Theory Level 4

Define a positive integer $n$ to be squarish if either $n$ is a square or the difference of $n$ to its nearest perfect square is a square. For example 2016 is a squarish since its difference from $45^2=2025$ is a square. Define $S(N)$ to be the number of squarish integers between 1 and $N$ inclusive. Find positive constants $\alpha$ and $\beta$ such that

$\large \lim_{N \to \infty} \frac {S(N)}{N^\alpha} = \beta$

Write the answer as $(\alpha, \beta)$ if it exists.

No such constants exist.

\left(\frac 43, \frac 34\right)

\left(\frac 12, \frac 23\right)

\left(\frac 34, \frac 43\right)

1 solution

Calvin Lin Staff
Oct 7, 2017

Hint: The number of squarish numbers from $k^2$ to $(k+1)^2$ is approximately $2\sqrt{k}$ . Hence the number of squarish number from $1$ to $k^2$ is approximately $\frac{4}{3} k ^ \frac{3}{2}$ . So the number of squaring numbers from $1$ to $n$ is approximately $\frac{4}{3} n^ \frac{3}{4}$ .

(Slight justification needed, but the idea is there.)

I think maybe you need to multiply all your estimates by $2$ ?

I don't remember the details, but when I was working the problem out I grouped the numbers by their nearest square, so there were about $2\sqrt{k}$ squarish numbers whose nearest square was $k^2.$

This is Putnam 2016 B-2, by the way.

Patrick Corn - 3 years, 7 months ago

Ah yes indeed. In that range, there are $\sqrt{k}$ perfect squares that "round" to $k^2$ , and another $\sqrt{k}$ perfect squares that round to $(k+1)^2$ .

If it's a Putnam problem, then there will need to be more justification in the solution, especially with the "Riemann sum". In particular, we need to sum $2 \lfloor \sqrt{k} \rfloor$ instead, and show that it is still approximate. But since $\lfloor \sqrt{k} \rfloor \sim \sqrt{k}$ , it all works out (just fill in the details).

Calvin Lin Staff - 3 years, 7 months ago

Squarish numbers

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