Möbius Inversion Formula

The Möbius function is defined as follows: ( \mu(n) = \begin{cases} 1 & \text{if } n \text{ is square-free} \ -1 & \text{if } n \text{ has an even number of prime factors} \ 0 & \text{otherwise} \end{cases} )

The Möbius Inversion Formula states that if ( f ) is a multiplicative function and ( g ) is defined by ( g(n) = \sum_{d|n} f(d) ), then ( f(n) = \sum_{d|n} \mu(d) g(\frac{n}{d}) ).

The Euler phi function is a multiplicative function, meaning that ( \phi(mn) = \phi(m) \phi(n) ) whenever ( m ) and ( n ) are relatively prime.

The sum-of-divisors function is an additive function, meaning that ( \sigma(mn) = \sigma(m) + \sigma(n) ) whenever ( m ) and ( n ) are relatively prime.

Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = 1 ).

Using the Möbius Inversion Formula with ( f(n) = d ) and ( g(n) = \sum_{d|n} \mu(d) d ), we get ( d = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \sigma(\frac{n}{d}) ) for all ( d | n ), we have ( \sum_{d|n} d = \sigma(n) ).

Using the Möbius Inversion Formula with ( f(n) = \phi(n) ) and ( g(n) = \sum_{d|n} \mu(d) \phi(d) ), we get ( \phi(n) = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \phi(\frac{n}{d}) ) for all ( d | n ), we have ( \sum_{d|n} \phi(d) = \phi(n) ).

Since ( n ) is square-free, all of its divisors are also square-free. Therefore, ( \mu(d) = 1 ) for all ( d | n ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = 1 ).

If ( n ) has exactly ( k ) prime factors, then all of its divisors have between 0 and ( k ) prime factors. Therefore, ( \mu(d) = (-1)^k ) for all ( d | n ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = (-1)^k ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } k \text{ is even} \ -1 & \text{if } k \text{ is odd} \end{cases} ).

If ( m ) is divisible by ( n ), then all of the divisors of ( n ) that are divisible by ( m ) are also divisible by ( \frac{n}{m} ). Therefore, ( \mu(d) = 0 ) for all ( d | n ) that are divisible by ( m ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 0 ) for all ( d | n ) that are divisible by ( m ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } m = 1 \ 0 & \text{if } m > 1 \end{cases} ).

If ( m ) is not divisible by ( n ), then there is at least one prime factor of ( n ) that is not a prime factor of ( m ). Therefore, there is at least one divisor of ( n ) that is not divisible by ( m ). Since ( \mu(d) = 0 ) for all ( d | n ) that are divisible by ( m ), we have ( \sum_{d|n} \mu(d) = -1 ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = -1 ) for all ( d | n ) that are not divisible by ( m ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 0 & \text{if } m = 1 \ -1 & \text{if } m > 1 \end{cases} ).

If ( d ) is a divisor of ( n ) that is relatively prime to ( n ), then ( \frac{n}{d} ) is also a divisor of ( n ) that is relatively prime to ( n ). Therefore, ( \mu(d) = \mu(\frac{n}{d}) ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \phi(\frac{n}{d}) ) for all ( d | n ) that are relatively prime to ( n ), we have ( \sum_{d|n} \mu(d) = \phi(n) ).

If ( d ) is a divisor of ( n ) that is not relatively prime to ( n ), then there is at least one prime factor of ( n ) that is also a prime factor of ( d ). Therefore, ( \mu(d) = 0 ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 0 ) for all ( d | n ) that are not relatively prime to ( n ), we have ( \sum_{d|n} \mu(d) = 0 ).

If ( n ) is a perfect square, then all of its prime factors appear with even exponents. Therefore, ( \mu(d) = 1 ) for all ( d | n ) that are perfect squares. Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ) that are perfect squares, we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } n \text{ is a perfect square} \ 0 & \text{otherwise} \end{cases} ).