# A Problem on the Möbius Function

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

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

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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.

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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.

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Could you please elaborate this? –  draks ... Jul 7 '12 at 22:10
@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