Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do we prove that $\Big[\sqrt n \Big]+ \sum_{j=1}^n \bigg[ \dfrac nj\bigg]$ is an even integer for all $ n \in \mathbb N$ ? (where $\Big[ \space \Big]$ denotes the "greatest integer" function)

share|improve this question
    
It looks like it is integer by definition... –  Bach Mar 15 at 14:06
    
@Bach: Yes , it is an integer ... –  Souvik Dey Mar 15 at 14:10
    
Sorry, I missed the even in your question. –  Bach Mar 15 at 14:13
1  

1 Answer 1

up vote 6 down vote accepted

$\left[\frac{n}{j}\right]$ is the number of positive integers $k$ such that $k\cdot j \leqslant n$. Thus

$$\sum_{j=1}^n \left[\frac{n}{j}\right] = \sum_{j\cdot k \leqslant n} 1 = \sum_{m = 1}^n \left(\sum_{j\cdot k = m} 1\right) = \sum_{m=1}^n \tau(m),$$

where $\tau(m)$ is the number of divisors of $m$. Every positive integer that is not a perfect square has an even number of divisors, while a perfect square has an odd number of divisors. There are $\left[\sqrt{n}\right]$ perfect squares $\leqslant n$. Hence

$$\sum_{j=1}^n \left[\frac{n}{j}\right] \equiv \left[\sqrt{n}\right] \pmod{2}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.