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.

$$\sum_{d^2|n}{\mu(d)}=|\mu(n)|$$

I don't know how to prove it, and could you give me some suggestions?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Each positive integer $n$ can be written uniquely in the form $$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$$ where $p_i$ are primes and $a_i$ are positive integers. Perform Euclidean division for each $a_i$ by $2$ to write $a_i=2 b_i+\epsilon_i$ so that $$n=p_1^{\epsilon_1}p_2^{\epsilon_2}...p_k^{\epsilon_k} x^2 $$ Notice that each $\epsilon_i$ is either 0 or 1 so you have written uniquely $n$ in the form $n=y x^2$ where $y$ is the largest squarefree integer which divides $n$ and $x^2$ is an integer square. Then the squares $d^2$ divide $n$ if and only if they divide $x^2$ which means that you really have the following sum $$\sum_{d|x}\mu(d)$$ This however equals 0 or 1, according to as $x>1$ or $x=1.$ The case $x=1$ takes place only when $n=y$ is squarefree and therefore the sum is 1 if $n$ is squarefree and 0 otherwise. This happens to coincide with $|\mu(n)|.$

You can use this type of Euclidean division in the exponents to derive similar identities for cube-free integers etc. You can also use your identity to detect how many squarefree integers there are below any number $t$:

Their cardinality equals $$\sum_{n\leq t} |\mu(n)|=\sum_{n \leq t}\sum_{d^2|n}\mu(d)=\sum_{d \leq \sqrt{t}}\mu(d)[\frac{t}{d^2}]=t(\frac{1}{\zeta(2)}+O(\frac{1}{\sqrt{t}}))=\frac{6}{\pi^2}t+O(\sqrt{t})$$ which means that the probability that a random integer $n$ is squarefree is equal to $\frac{6}{\pi^2}.$ As this number is more than one half, you get for free the Goldbach-like statement, that each integer can be written as the sum of two squarefree numbers.

share|improve this answer
1  
Thanks for your great answer! –  Kou Nov 16 '11 at 3:55
    
Indeed. Just what I was looking for. –  draks ... Jul 7 '12 at 22:07

If you can prove that each side is a multiplicative function, then you only have to prove the formula for $n$ a power of a prime.

share|improve this answer
    
Could you please elaborate this? –  draks ... Jul 7 '12 at 22:10
1  
@draks, $f$ is multiplicative means if $\gcd(m,n)=1$ then $f(mn)=f(m)f(n)$. Since every positive integer can be written as a product of prime powers, and since powers of distinct primes are relatively prime, it follows that if $f$ is multiplicative and you know $f(p^k)$ for all prime powers $p^k$ then you get a formula for $f(n)$ for all $n$. –  Gerry Myerson Jul 8 '12 at 3:31
    
Thanks ${}{}{}{}$ –  draks ... Jul 8 '12 at 10:25

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.