Möbius Inversion Formula and Applications to Cyclotomic Polynomials

Degree Project Möbius Inversion Formula and Applications to Cyclotomic Polynomials 2012-06-01 Author: Zeynep Islek Subject: Mathematics Level: Bachelor Course code: 2MA11E

Abstract This report investigates some properties of arithmetic functions. We will prove Möbius inversion formula which is very important in number theory. Also our report investigates roots of unity and cyclotomic polynomials over the complex numbers. 2

Contents 1 Introduction 4 2 Arithmetic Functions 7 2.1 The Möbius Function...................... 7 2.2 The Euler Function........................ 10 2.3 Möbius Inversion Formula.................... 13 3 Cyclotomic Polynomials and Roots of Unity 16 3.1 nth Root of Unity........................ 16 3.2 Cyclotomic Polynomials..................... 18 4 Conclusion 23 3

1 Introduction The number theory is a very important part of mathematics. We can say that it is a basis of mathematics. Carl Freidrich Gauss who is a very famous mathematician, was said that : Mathematics is the queen of science and arithmetics is the queen of mathematics. The classic Möbius function is an important multiplicative function in the number theory and combinatorics. In number theory there are two very important multiplicative functions which are Möbius function and Euler s function, denoted by µ and ϕ respectively. The Möbius µ introduced to solve a problem which was related to Riemann zeta function. The Euler s function ϕ introduced to generalize a congruence result of Fermat but in this study we are not interested in congruence relating and Riemann zeta function. The classic Möbius function is defined by the August Ferdinant Möbius, in 1832. It is a function whose domain is the positive integers, and which is defined as follows: µ(n) = 1 if n = 1 0 if n is divisible by a square bigger than 1 ( 1) k if n is product of k distinct primes Dedekind and Liouville reported the inversion theorem for sums, simultaneously. In 1857 they gave some appliction to ϕ(m). R.Dedekind redeploid the function in the reverse of series which is given by Möbius. E.Laguerre described the function below, using the function which is written by Dedekind. If F (m) = f(d) where d ranges over the divisors of m, then f(m) = µ( m )F (d). (1) d d m When these formulas and (1) were used, this formula was gotten: ϕ(d) = m. F. Mertens noted that if n > 1, µ(d) = 0 where d ranges over the divisors of n. A.F. Möbius recognized its arithmetical importance, in 1832. Möbius anaylsed the inverse of f which is an arbitrary function, using the Dirichlet series. Liouville and Dedekind gave the finite form of the Möbius inversion formula, in 1857, as follows g(n) = f(d) f(n) = µ(d)g( n d ). 4

If you want to learn more details about history of these functions, please read [1]. Now we will mention relationship between Möbius function and roots of unity, but before we will give some definitions. A polynomial, not identically zero, is said to be irreducible if it cannot be written as a product of two or more non-trivial polynomials whose coefficients are of specified type. If you want to learn more, please check [2, 3, 4]. Every non-zero polynomial over C can be factored as p(x) = α(x z 1 )... (x z n ) where n is the degree, α is the leading coefficient and z 1,..., z n the zeroes of p(x). If α C and rational numbers c 1,..., c n exist satifying α n + c 1 α n 1 + + c n = 0 then α is called an algebraic numbers. If you want to read more details, please check [3, 5, 6, 7]. Let s take any algebraic number α. The minimal polynomial of α is the unique irreducible polynomial of the smallest degree p(x) with rational coefficients such that p(α) = 0 and whose leading coefficients 1. If you want to learn more details about minimal polynomials, please read [8]. In mathematics, a root of unity, is any complex number that equals 1 when raised to some integer power n. Roots of unity is important in number theory. An nth root of unity, is a complex number z satisfying the equation z n = 1 where n = 1, 2, 3,... is a positive interger. An each element of this sum, which is shown e 2πik n implies that nth root and k for the kth power. Equivalentely, we can use (e 2πi n ) k n instead of e 2πik n, in the complex plane. An nth root of unity is primitive if it is not a kth root of unity for some smaller k: z k 1, k = 1, 2,..., n 1. The zeroes of the polynomial p(z) = z n 1 are the nth roots of unity, each with multiplicity 1. There is a unique monic polynomial Φ n (x) having degree ϕ(n) whose root are the distinct primitive nth roots of unity, where ϕ is an Euler s function. Φ is called a cyclotomic polynomial. Finally, we can say that about this study. In the first part of this thesis, we emphasize the Möbius function and we prove the Möbius inversion formula. Using this Möbius inversion formula, we prove some theorems. In 5

the second part we emphasize the roots of unity in the complex numbers, correspondingly we emphasize the cyclotomic polynomials in the complex numbers, and we will see the connection between the Möbius inversion formula and the cyclotomic polynomials in the complex numbers. 6

2 Arithmetic Functions In this section we describe Möbius function and Möbius inversion formula, then we prove these functions. Also we prove some theorems which we need for proving Möbius inversion formula. Now we start to give some definition about the number theory. Definition 2.1. A real or complex valued function defined on the positive integers is called an arithmetic function. In set notation: f : Z + R or f : Z + C If you want to see more details on arithmetic functions, check [9]. Definition 2.2. An arithmetic function f is called multiplicative if f(mn) = f(m)f(n) where m and n are relatively prime positive integers (i.e. (m, n) = 1). Definition 2.3. An arithmetic function f is called completely multiplicative if f(mn) = f(m)f(n) for every positive integers m and n. Example 1. The function f(x) = 1 x arithmetic function where f : Z+ R, because since all x in Z +, the results are in R. Let s take x = 2, then f(x) = 1 x = 1 2, and 1 2 is in R. Now it is necessary to learn next definition for continue apprehensibly. Definition 2.4. Let a, b Z and a 0 such that b = ax if there exist x Z, then we say that a divides b which can be denoted a b, and a b if and only if b = ax for all x in Z. 2.1 The Möbius Function The arithmetic function µ(n), defined for all natural numbers, is called Möbius function. Definition 2.5. The Möbius function µ(n) is defined as follows 1 if n = 1 µ(n) = 0 if n is divisible by a square larger than 1 ( 1) k if n = p 1... p k where p i s are relatively prime numbers 7

Example 2. We have µ(1) = 1 it is clear to see from the definition. Moreover µ(2) = 1 because 2 is a prime number. So µ(2) = ( 1) 1 = 1. We have µ(4) = 0 because 2 2 4 or we can say that 4 is divisible by a square. We have because 2 2 8. We have µ(8) = 0 µ(42) = 1 because 42 = 2 3 7, so 42 can be written as the product of three relatively prime numbers. Thus µ(42) = ( 1) 3 = 1. Theorem 2.1. The function µ(n) is multiplicative. Proof. We will prove that µ(mn) = µ(m)µ(n) whenever m and n are relatively prime numbers. First, we consider m and n are square-free numbers. We assume that m = p 1... p k, where p 1,..., p k are distinct primes, and n = q 1... q s, where q 1,..., q s are distinct primes. From the definition of µ(n), we write that µ(m) = ( 1) k and µ(n) = ( 1) s, and mn = p 1... p k q 1... q s, again using the definition of µ(n), we write µ(mn) = ( 1) k+s. Hence µ(mn) = ( 1) k+s = ( 1) k ( 1) s = µ(m)µ(n). Now suppose at least one of m and n is divisible by a square of a prime, then mn is also divisible by the square of a prime. So µ(mn) = 0 and µ(m) or µ(n) is equal to zero. Now it is clear to see that the product of µ(m) and µ(n) is equal to zero. So µ(mn) = µ(m)µ(n). On the other hand, from the definition of µ(n), we know that µ(4) = 0 because 2 2 4 and µ(2) = 1. We can write that µ(4) = µ(2 2), but µ(4) = 0 µ(2)µ(2). Hence µ(n) is not completely multiplicative function. The Möbius function appears in many different places in number theory. One of its the most important properties is a formula for the divisor sum µ(d), extended over the positive divisor of n. It leads to Möbius inversion formula. 8

Theorem 2.2. For the Möbius function µ(n), the summatory function is defined by { 1 if n = 1 µ(d) = 0 if n > 1. We need the following theorem to prove Theorem 2.2. Theorem 2.3. If f is multiplicative function of n, and F is defined as follows F (n) = f(d) then F is also multiplicative function. Proof. We will show that F is multiplicative function. If F is multiplicative function, we write that when m and n are relatively numbers, then F (mn) = F (m)f (n). So, now let us choose (m, n) = 1. We have F (mn) = d mn f(d). Now, all divisiors of mn must be written as the product of relatively prime numbers. As we mentioned before, if F is multiplicative function, we write F (mn) = F (m)f (n) when (m, n) = 1. So we write d = d 1 d 2 as the product of relatively prime divisors d 1 of m and d 2 of n. Hence, we write F (mn) = d 1 m d 2 n f(d 1 d 2 ). Since f is multiplicative and since (d 1, d 2 ) = 1, we can write that F (mn) = d 1 m d 2 n f(d 1 )f(d 2 ) = d 1 m f(d 1 ) d 2 n f(d 2 ) = F (m)f (n). Now we continue to prove the Theorem 2.2. Proof. Consider, n = 1. It is clear to see that µ(d) = µ(d) = µ(1) = 1. d 1 9

Now we assume this formula for n > 1. Let us define an arithmetic function M as M(n) = µ(d). The Möbius function is multiplicative, then M(n) is multiplicative by Theorem 2.3. Let s suppose that n which is the product of powers of r different relatively prime numbers such that n = r i=1 p a i i. Then the results which are under the function of M are equal. So we write M(n) = r i=1 M(p a i i ). Now we are searching what is the result of M(n). If we find the result of M(p a i i ), we find the result of M(n). Now we can write that M(pa i i ) = d p a i µ(d) using the Theorem 2.3, which is the way to find the result. We i have M(p a i i ) = µ(d) d p a i i = µ(1) + µ(p i ) + µ(p 2 i ) + + µ(p a i i ) = 1 + ( 1) + 0 + + 0 = 0. For every integer bigger than 1, we proved that the sum function of Möbius function is equal to zero. The another example of an arithmetic function is Euler s function. It is also a multiplicative function. 2.2 The Euler Function The function was introduced by Euler, in 1760, and is denoted by ϕ. This function is multiplicative which is one of the most important function in number theory. Definition 2.6. The Euler function ϕ(n) is the number of positive integers less than n which are relatively prime to n. 10

Example 3. Here is some values of ϕ(n). ϕ(1) = 1, ϕ(2) = 1, ϕ(3) = 2, ϕ(4) = 2, ϕ(5) = 4 ϕ(6) = 2, ϕ(7) = 6, ϕ(8) = 4, ϕ(9) = 6, ϕ(10) = 4 Now we find the values of the phi-function at primes powers. Theorem 2.4. Let p be a prime number. Then ϕ(p α ) = p α p α 1. Proof. Between 1 and p α there are p α integers. There are some numbers which are not relatively prime to p α, they are p, 2p,..., p α 1. There are exactly p α 1 such integers. So there are p α p α 1 integers less than p α that are relatively prime to p α. Hence, ϕ(p α ) = p α p α 1. Theorem 2.5. If p is a prime, then ϕ(p) = p 1. Proof. It is easy to see that from Theorem 2.4. Now we suppose α = 1 then ϕ(p α ) = p α p α 1 = p 1 p 1 1 = p 1. Example 4. We calculate the values of ϕ(n) for some prime numbers. ϕ(5 3 ) = 5 3 5 2 = 100 ϕ(2 10 ) = 2 10 2 9 = 512 ϕ(11 2 ) = 11 2 11 1 = 110 The Euler s ϕ function is multiplicative function. If you re interested in proof of ϕ is multiplicative function, please read [3, 5, 7]. Example 5. Let s calculate ϕ(756) using that the Euler s ϕ function is multiplicative. This number can ben be written as 756 = 2 2 3 3 7. Hence ϕ(756) = ϕ(2 2 3 3 7). We know that ϕ function is multiplicative. So we write ϕ(756) = ϕ(2 2 ) ϕ(3 3 ) ϕ(7). Using the Theorem 2.4 and Theorem 2.5, we find ϕ(2 2 ) = 2 2 2 = 2, 11

So ϕ(3 3 ) = 3 3 3 2 = 18, ϕ(7) = 7 1 = 6. ϕ(756) = 2 18 6 = 216. Now we get the result. If you want to check more examples, read [5]. Theorem 2.6. For every positive integers d and n, we have ϕ(d) = n. Proof. We will prove this by induction on the number of different prime factors. We consider the case n = p α, where p is a prime number. We have ϕ(d) = ϕ(d) = ϕ(1) + ϕ(p) + ϕ(p 2 ) + + ϕ(p α ) d p α = 1 + (p 1) + (p 2 p) + + (p α p α 1 ) = p α = n. Correspondingly, now suppose that the theorem holds for integers with k distinct prime factors. Let us take any integer N with k +1 distinct prime factors and p α be the highest power of p that divides N. Now we write N = p α n where p and n are relatively prime numbers (i.e., (p,n)=1). We know when d ranges over the divisor of n, the set d, dp, dp 2,..., dp α ranges over the divisors of N. Then d N ϕ(d) = = ϕ(d) + ϕ(d) + ϕ(dp) + ϕ(d)ϕ(p) + ϕ(dp 2 ) + + ϕ(dp α ) ϕ(d)ϕ(p 2 ) + + ϕ(d)ϕ(p α ) = ϕ(d)[1 + ϕ(p) + ϕ(p 2 ) + + ϕ(p α )] = = np α = N. ϕ(d) e p α ϕ(e) We showed that ϕ(d) = n is true for every positive integers n, and d ranges over n. 12

Example 6. We give an example to understand the Theorem 2.6. ϕ(d) = ϕ(1) + ϕ(2) + ϕ(3) + ϕ(4) + ϕ(6) + ϕ(12) d 12 = 1 + 1 + 2 + 2 + 2 + 4 = 12. REMARK: As ϕ is an arithmetic function, we note that ϕ(d) = ϕ( n d ). Now we show that for any multiplicative functions: If f is multiplicative function and not equal to zero, then f(d) or, equivalently, f(n d) denotes the sum of the values of a function f where d ranges over the positive divisors of n. We write that f(d) = f(n d) because since d ranges over n, n d ranges over n. For example; f(d) = f(1) + f(2) + f(3) + f(6) + f(9) + f(18) d 18 d 18 f( n d ) = f(18 1 ) + f(18 2 ) + f(18 3 ) + f(18 6 ) + f(18 9 ) + f(18 18 ) = f(18) + f(9) + f(6) + f(3) + f(2) + f(1) 2.3 Möbius Inversion Formula Theorem 2.7. If g is any arithmetic function and f is the sum function of g, so that f(n) = g(d) then g(n) = f(d)µ( n d ). Equivalently, if d ranges over n, n d ranges over n. Hence, we can write f(d)µ( n d ) = f( n d )µ(d). 13

Proof. The equality f(d)µ( n d ) = f( n d )µ(d) is true from the remark. If, we write n = ed from the definition (divisibility), where e is in Z. Now let us take n = de, so e = n d. Then the previous sum can be written as f(d)µ(e) de=n and it is possible to write the last sum as, f(e)µ(d). de=n Now we must prove that the sum f(d)µ( n d ) is equal to g(n) or, equivalentely, the sum f(n d )µ(d) is equal to g(n). Using equality below we write that f( n d ) = e n d g(e) µ(d)f( n d ) = (µ(d) e n d g(e)). Since e divides n d, then e divides n. Inversely, each divisor of n is e which divides n d if and only if d divides n e. So d divides n. As have seen, the coefficent of g(e) is d n µ(n) can be written as e µ(n) = d n e { 1 if n e = 1 0 if n e > 1 using the Theorem 2.2. That implies g(n) has only one coefficient g(e) which is not equal to zero. So g(e) = 1. Then g(n) = f( n d )µ(d). The Euler function is related to the Möbius function through the following formula. 14

Theorem 2.8. ϕ(n) = n µ(d) d where ϕ(n) is Euler s ϕ function. Proof. We know ϕ(d) = n from the Theorem 2.6. Take a function F which is the sum of the Euler s ϕ function is as F (n) = ϕ(d) = n. Use the Möbius inversion formula here ϕ(n) = F (d)µ( n d ) = = F ( n d )µ(d) n d µ(d) = n µ(d) d. Example 7. Let s calculate the value of ϕ(756), using Theorem 2.8. The divisors of this number are 1, 2, 3, 4, 7, 9, 12, 14, 18, 21, 27, 28, 36, 42, 54, 63, 84, 108, 126, 189, 252, 378, and 756. Now using the Theorem 2.8 we can calculate ϕ(756). ϕ(756) = n µ(d) d = 756 d 756 µ(d) d And here is some values of µ(n). = n d µ(d) ( µ(1) = 756 + µ(2) + + µ(378) + µ(756) ) 1 2 378 756 µ(1) = 1, µ(2) = 1, µ(3) = 1, µ(6) = 1, 15

µ(7) = 1, µ(14) = 1, µ(21) = 1, µ(42) = 1. The value µ of the other numbers are equal to zero, because these numbers are divisible by a square larger than 1. Hence ( 1 756 1 + 1 2 + 1 3 + 1 6 + 1 7 + 1 14 + 1 21 + 1 ) = 756 2 42 7 = 216 We solved same example using by ϕ function is a multiplicative function. That way is easy and short, because if you want to use the Theorem 2.8, you should know that the values of µ function. 3 Cyclotomic Polynomials and Roots of Unity 3.1 nth Root of Unity Assume that a be an nth root of a number b, this means that b n = a. In particular the square root of 1 is 1 because 1 1 = 1, but ( 1)( 1) = 1, so 1 also a square root of 1. There are two square roots of 1. If we take the cube root of 1, then 1 is not a solution because ( 1)( 1)( 1) = 1. So 1 has only one solution if we study on real numbers, but in the complex numbers, there are three roots of 1. All cube roots of 1 can also be defined as powers of the negative interval. We see them below 16

REMARK: Let n be an integer and x be a complex number (and, in particular, a real number), the Euler s function states that e i(nx) = cos(nx) + isin(nx). C be the field of complex numbers, there are exactly n different nth roots of 1. If you divide the unit circle into n equal parts, using n points, it is easy to find them. Definition 3.1. A complex number z is called an nth root of unity for a positive integer n, if z n = 1. We will show that z n = 1. Let s take any complex number z. If we write this complex number on the polar coordinates, we get If we take the nth power of z, z = cosθ + isinθ. z n = (cosθ + isinθ) n = cos(n θ) + isin(n θ). We can write this equality, because this is de Moivre s formula. So z n = 1 cos(n θ) + isin(n θ) = cos(0) + isin(0). 17

Then nθ = 0 + 2kπ, where k = 0, 1,..., n 1. Thus θ = 2kπ n. The roots of unity is then in e 2πi/n for k = 0, 1,..., n 1. There are n different solutions for z n = 1, namely, e 2πi/n, e 2πi2/n,..., e 2πin/n. We usually assume that ζ n = e 2πi/n so that ζ n, ζ 2 n,..., ζ n n are the nth roots of z n = 1. Definition 3.2. An nth root of unity is primitive if it is of the form ζ k n with k and n relatively prime numbers, i.e., (k, n) = 1. If ζ n is a primitive nth root of unity and (ζ k n) m = 1 then n m. If you want to read more details, check [6]. 3.2 Cyclotomic Polynomials Definition 3.3. Let n be a positive integer and let ζn k be the primitive nth root of unity ( ζ n is the complex number e 2πi/n ). The nth cyclotomic polynomial Φ n (x) is Φ n (x) = (x ζn) k 1 k n gcd(n,k)=1 whose roots are the primitive nth roots of unity. Theorem 3.1. Let n be a positive integer. Then x n 1 = Φ d (x) where d ranges over the divisor of n. Proof. The roots of x n 1 are exactly nth roots of unity. On the other hand, if ζ is an nth root of unity and the order of ζ is d, then ζ is a primitive dth root of unity. So ζ is a root of Φ d (x). But, so ζ is a root of the right hand side. It follows that the polynomials on the left and right hand side have the same roots. Thus they are equal. Another way to find the nth cyclotomic polynomial: If n > 1, then Φ n (x) = xn 1 d Φ d(x) where d ranges over, except n, the divisor of n. (2) If you want to see more detail about cyclotomic polynomials, check [4, 10]. 18

Example 8. Here is some value of cyclotomic polynomials. Φ 1 (x) = x 1 Φ 2 (x) = x2 1 Φ 1 (x) = x2 1 x 1 = x + 1 Φ 3 (x) = x3 1 Φ 1 (x) = x3 1 x 1 = x2 + x + 1 Φ 4 (x) = x4 1 Φ 1 (x)φ 2 (x) = x2 + 1 Φ 5 (x) = x5 1 Φ 1 (x) = x4 + x 3 + x 2 + x + 1 Φ 6 (x) = x 6 1 Φ 1 (x)φ 2 (x)φ 3 (x) = x2 x + 1 Now we know that x n 1 = Φ d(x), and conversely, by using the Möbius function, we can write the following theorem. Theorem 3.2. Let n be a positive integer and µ(n) denotes the Möbius function. Then Φ n (x) = (x d 1) µ( n d ). Proof. To prove this formula, first we use the equality x n 1 = Φ d(x), then we take complex logarithm of this equality, and finally use the Möbius inversion formula. We have x n 1 = Φ d (x). Now take the complex logarithm both side of equality. It doesn t effect to the equality. log(x n 1) = log Φ d (x) Let s assume d = {d 1, d 2,..., d s }, where d i s are divisor of n and are not equal to n. log Φ d (x) = log (Φ d1 (x) Φ d2 (x) Φ ds (x)) 19

Using the property of logarithm, we write log (Φ d1 (x) Φ d2 (x) Φ ds (x)) = log(φ d1 (x)) + log(φ d2 (x)) + + log(φ ds (x)) = log(φ d (x)). Using Möbius inversion formula, we can write log(φ n (x)) = µ( n ) (log(x d ) d 1) = (log(x d 1) µ( n )) d = log((x d 1 1) µ( n ) d 1 ) + log((x d 2 1) µ( n ) d 2 ) + + log((x d s 1) µ( n ds ) ) = log ((x d 1 1) µ( n ) d 1 (x d 2 1) µ( n ) d 2 (x d s 1) µ( n )) ds = log (x d 1) µ( n d ) If we cancel out the logarithm, we get the result Φ n (x) = (x d 1) µ( n d ). Now we can see the connection between Möbius inversion formula and cyclotomic polynomials. When x n 1 = Φ d (x), it is possible to write that Φ n (x) = (x d 1) µ( n d ). Example 9. We are going to find cyclotomic polynomials using the formula Φ n (x) = (x d 1) µ( n d ) for n = 1, 2,..., 20. Φ 1 (x) = x 1 Φ 2 (x) = x + 1 Φ 3 (x) = x 2 + x + 1 20

Φ 4 (x) = x 2 + 1 Φ 5 (x) = x 4 + x 3 + x 2 + x + 1 Φ 6 (x) = x 2 x + 1 Φ 7 (x) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 8 (x) = x 4 + 1 Φ 9 (x) = x 6 + x 3 + 1 Φ 10 (x) = x 4 x 3 + x 2 x + 1 Φ 11 (x) = x 10 + x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 12 (x) = x 4 x 2 + 1 Φ 13 (x) = x 12 + x 11 + x 10 + x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 14 (x) = x 6 x 5 + x 4 x 3 + x 2 x + 1 Φ 15 (x) = x 8 x 7 + x 5 x 4 + x 3 x + 1 Φ 16 (x) = x 8 1 Φ 17 (x) = x 16 + x 15 + x 14 + x 13 + x 12 + x 11 + x 10 + x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 18 (x) = x 6 x 3 + 1 Φ 19 (x) = x 18 + x 17 + x 16 + x 15 + x 14 + x 13 + x 12 + x 11 + x 10 + x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 20 (x) = x 8 x 6 + x 4 x 2 + 1 We have already found some cyclotomic polynomials in example 8. In this example you don t need to know values of Möbius function. If you want to find cyclotomic polynomials using the Theorem 3.2, you should know the values of Möbius function. As have seen, the coefficients of cyclotomic polynomial are often 1, 0 and 1, but for n 105 some coefficients are different from this set. For example n = 105, then Φ 105 (x) =x 48 + x 47 + x 46 x 43 x 42 2x 41 x 40 x 39 + x 36 + x 35 + x 34 + x 33 + x 32 + x 31 x 28 x 26 x 24 x 22 x 20 + x 17 + x 16 + x 15 + x 14 + x 13 + x 12 x 9 x 8 2x 7 x 6 x 5 + x 2 + x + 1. We see that there is a coefficient 2 which is not included in { 1, 0, 1}. Theorem 3.3. The coefficients of Φ n (x) are integers. Proof. We prove it using inductive method. Clearly, Φ 1 (x) = x 1 Z[x]. Now we suppose that, for k < n, the coefficients of Φ n (x) are integers. Let f(x) = Φ d (x). Then we say that f(x) Z[x], from the inductive d<n method. Using the formula (2), we write if Φ n (x) = xn 1 f(x) 21

then x n 1 = Φ n (x)f(x) On the other hand, x n 1 Z[x]. Also using the division algorithm, we can write x n 1 = f(x)g(x) + r(x) for some g(x), r(x) Z[x]. By the uniqueness, we take r(x) = 0. So x n 1 = f(x)g(x). It is easy to see g(x) = Φ n (x). Since g(x) Z[x], Φ n (x) Z[x]. Finally we say that the coefficients of Φ n (x) are integers. 22

4 Conclusion In this study, the Möbius inversion formula has been introduced and proved. The roots of unity and cyclotomic polynomials have been introduced. In the first cheapter, some informations about Möbius function, Euler s ϕ function, Möbius inversion formula, roots of unity and cyclotomic polynomials have been introduced. And also some definitions which will be used in the next cheapters have been mentioned. In the second cheapter, some arithmetic functions and properties of these functions have been studied. The Möbius function which is defined on arithmetic functions, has been studied. The Euler s ϕ function and its applications have been mentioned. The Möbius inversion formula has been introduced. A theorem which is related with Euler s ϕ function and µ function, using Möbius inversion formula, has been proved, that is ϕ(n) = n µ(d) d. In the third cheapter, the roots of unity in the complex numbers have been introduced. Accordingly, the primitive roots of unity have been introduced. Later, the cyclotomic polynomials whose roots are the primitive nth roots of unity, have been investigated. Accordingly, the connection between Möbius inversion formula and cyclotomic polynomials has been shown. The connection between Möbius inversion formula and cyclotomic polynomials in the complex numbers has been shown. Since this is the formula below, x n 1 = Φ d (x). and with the Möbius inversion formula and using the complex logarithm Φ n (x) = (x d 1) µ( n d ) has been proved. 23

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