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n-th non-square

Problem:
Find the n-th non-square number.

(Adapted from Problem Solving Strategies, Arthur Engel)


If you have CS background, I invite you to come up with an O(log N) solution.

Done? Good. Below I present the O(1) approach.

Solution:
The n-th non-square number is given by \( F(n) = \left \lfloor n + \sqrt{n} + 1/2  \right \rfloor \).

Proof:
Let G(n) be the same as F(n) except for the floor function.

Investigate \( m \neq F(n) \). There is the largest \( n \) s.t. \( F(n) < m \). Hence \( m+1 \leq F(n+1) \).
We conclude that \(G(n) < m \) and \(m+1 \leq G(n+1) \).

Rearranging the terms and squaring the inequalities, we get:
\( n < (m-n-1/2)^2 \) and \( (m-n-1/2)^2 < n+1 \), or
\( n - 1/4 < (m-n)^2 - (m-n) < n + 3/4 \).
Since the middle term is an integer, we see that \( (m-n)^2 - (m-n) = n \). Simplifying, we get \(m = (m-n)^2\), showing that m is a square.

In other words, F(n) skips all squares.

Now, we will show that F(n) is the n-th non-square number. The proof is by counting.
Fix k. We note that \(F(k^2) = k^2 + k\). Hence There are \(k^2\) terms of F(n) \( \leq k^2 + k \).
On the other hand, we know that there are k squares \( \leq k^2 + k \). Hence F(n) iterates through all the non-square numbers \( \leq k^2 + k \). Therefore we conclude that F(n) is the n-th non-square.

QED

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