Skip to main content

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

Comments

Popular posts from this blog

641 | 2^32 + 1

Problem: Show that \( 641 | 2^{32} + 1 \). Solution: (From Problem Solving Strategies, Arthur Engel) \( 641 = 625 + 16 = 5^4 + 2^4 \). So \( 641 | 2^{32} + 2^{28} \cdot 5^4 \). Also, \( 641 = 640 + 1 = 2^7 \cdot 5 + 1\). So \( 641 | (2^7 \cdot 5)^4 - 1 = 2^{28}\cdot 5^4 - 1 \). Hence \( 641 | 2^{32} + 2^{28} \cdot 5^4 -(2^{28}\cdot 5^4 - 1) \). QED

Derive Poisson Distribution

Let's say we are observing an event that to us seems random. E.g. The number of cars passing through a road. Suppose we learn that on average, there are m cars passing the road within 1 hour. Goal: Can we make a model to suggest what is the probability of seeing exactly r cars passing the road in an hour? Let's represent one hour as a line, and on the line there are marks to represent the time on which a car passes. Let's suppose there are m marks, to represent the average case of seeing m cars on the road in an hour. Let's divide the 1 hour line into n uniform time segments, n large enough such that in one time segment there is at most 1 car. Out of the n time segments, m of them contains a car. Then it is reasonable to say that the probability of a time segment to contain a car is m/n. Let's analyse as n goes to infinity. We can say P(seeing exactly r cars in an hour) =  P(exactly r cars contained in n segments) = \( \binom{n}{r} \left( \frac{m}{n} \right)^r \left...