Jay Daigle Occidental College Math 322: Number Theory 5 Multiplicative Functions For further reading on the material in this subsection, consult Rosen 7.1, Stein 2.2. Definition 5.1. An arithmetic function is a function defined for all natural numbers. A function is multiplicative if it has the property that f(mn) = f(m)f(n) whenever (m; n) = 1. It is completely multiplicative if f(mn) = f(m)f(n) for all natural numbers m; n. Example 5.2. The functions f(n) = 1 and g(n) = n are completely multiplicative. We will see that φ(n) is multiplicative but not completely multiplicative. (Example: φ(4) = 2 6= 1 · 1 = φ(2) · φ(2)). Completely multiplicative functions are easy to understand, but we can get a good grasp even of regularly multiplicative functions. a1 a2 as Qs ai Proposition 5.3. If f is multiplicative and n = p1 p2 : : : ps = i=1 pi is the prime factorization of n, then s a1 a2 as Y ai f(n) = f(p1 )f(p2 ) : : : f(ps ) = f(pi ): i=1 Proof. We prove by induction on s, the number of distinct prime factors of n. If s = 1 then a1 a1 n = p1 and then f(n) = f(p1 ) is trivially true. Suppose the proposition is true for all integers with k distinct prime factors, and suppose Qk+1 ai Qk ai ak+1 n has k + 1 distinct prime factors, say n = i=1 pi . We observe that i=1 pi ; pk+1 = 1, and thus by definition of a multiplicative function we know that k+1 ! k ! Y ai Y ai ak+1 f pi = f pi f(pk+1 ): i=1 i=1 And by inductive hypothesis we know that k ! k Y ai Y ai f pi = f(pi ) i=1 i=1 and thus we have k+1 ! k k+1 Y ai Y ai ak+1 Y ai f pi = f(pi )f(pk+1 ) = f(pi ): i=1 i=1 i=1 Thus for any multiplicative function, if we can compute its value for prime powers, we can easily compute its value for any number. http://jaydaigle.net/teaching/courses/2016-fall-322/ 57 Jay Daigle Occidental College Math 322: Number Theory 5.1 The Euler φ-function For further reading on the material in this subsection, consult Rosen 7.1, Stein 2.2. We want to understand the Euler φ-function much better. We will prove that it is a multiplicative function (though, as we observed earlier, it is not completely multiplicative). After that we'll figure out how to compute φ of prime powers, which will allow us to easily compute φ(n) for any positive integer n. Proposition 5.4. Let m; n be relative prime natural numbers. Then φ(mn) = φ(m)φ(n). In other words, the function φ(n) is multiplicative. Proof. Write the numbers ≤ mn as follows: 1 m + 1 2m + 1 ::: (n − 1)m + 1 2 m + 2 2m + 2 ::: (n − 1)m + 2 3 m + 3 2m + 3 ::: (n − 1)m + 3 . .. r m + r 2m + r : : : (n − 1)m + r . .. m 2m 3m : : : nm Note that (km + r; m) = (r; m), thus the first element of a given row is relatively prime to m if and only if every element of that row is. Since (r; m) > 1 implies that (r; mn) > 1, we only need to consider elements in rows r where (r; m) = 1. There are, of course, φ(m) such rows. Now suppose (r; m) = 1 and consider the elements of this row, which are km + r for 0 ≤ k ≤ n − 1. We claim this is a complete system of residues modulo n. It's enough to prove that no two elements are congruent to each other, by HW 4 problem 2. But if im + r ≡ jm + r mod n then njm(i − j), and since (n; m) = 1, by Euclid's lemma this implies nji − j. But i; j < n, so i = j. Since this is a complete system of residues, exactly φ(n) of these integers are relatively prime to n. Since these integers are also relatively prime to m, they are relatively prime to mn. Thus there are φ(m) rows that contain any elements relatively prime to mn; each such row contains φ(n) such elements. Thus there are in total φ(m)φ(n) natural numbers relatively prime to mn and ≤ mn; but this is the definition of φ(mn). http://jaydaigle.net/teaching/courses/2016-fall-322/ 58 Jay Daigle Occidental College Math 322: Number Theory Now that we know φ(n) is a multiplicative function, we know we can compute it purely by computing its value at prime powers. So we turn our attention to computing φ(pk). First, recall from homework that φ(n) = n − 1 if and only if n is prime. Lemma 5.5. Let p be a prime number, and let k be a positive integer. Then φ(pk) = pk−pk−1. Proof. An integer is relatively prime to pk if and only if it is divisible by p. Thus the integers n ≤ pk which are not relatively prime to pk are the integers `p for 1 ≤ ` ≤ pk−1. There are of course pk−1 such integers, and there are pk total integers n ≤ pk; thus there are pk − pk−1 integers n ≤ pk such that (n; pk) = 1. Example 5.6. φ(210) = 210 − 29 = 1024 − 512 = 512: φ(73) = 73 − 72 = 343 − 49 = 298. Qk ai Theorem 5.7. Let n = i=1 pi be the prime factorization of a natural number. Then k 1 1 1 Y 1 φ(n) = n 1 − 1 − ::: 1 − = n 1 − : p p p p 1 2 k i=1 i Proof. By proposition 5.3, we know that k Y ai φ(n) = φ(pi ): i=1 But by lemma 5.5 we know that ai ai ai−1 ai 1 φ(pi ) = pi − pi = pi 1 − : pi Thus k k Y Y 1 φ(n) = φ(pai ) = pai 1 − i i p i=1 i=1 i k ! k ! Y Y 1 = pai 1 − i p i=1 i=1 i k Y 1 = n 1 − : p i=1 i Example 5.8. 1 4 φ(100) = φ(22 · 52) = 100(1 − 1=2)(1 − 1=5) = 100 · · = 40: 2 5 8 φ(360) = φ(23 · 32 · 5) = 360(1 − 1=2)(1 − 1=3)(1 − 1=5) = 360 = 96: 30 http://jaydaigle.net/teaching/courses/2016-fall-322/ 59 Jay Daigle Occidental College Math 322: Number Theory Corollary 5.9. If n > 2 then φ(n) is even. Qk ai Qk ai Proof. Let n = i=1 pi . Then φ(n) = i=1 φ(pi ): ak ak−1 ak Suppose n has an odd prime factor pk. Then since pk and pk are both odd, 2jpk − ak−1 pk = φ(pk)jφ(n). Now suppose n has no odd prime factors. Then n = 2r and r > 1. Then φ(n) = 2r − 2r−1 = 2r−1 is even. This opens up an additional question: given an integer m, for what n is φ(n) = m? Example 5.10. What are the solutions to the equation φ(n) = 8? a1 a2 ak Suppose n = p1 p2 : : : pk . Then we have the equation k Y aj −1 φ(n) = pj (pj − 1) j=1 which is just a restatement of theorem 5.7. Then the only primes that can divide n are 2, 3, and 5, since we know that pi − 1jn. Further, if ai > 1 then pijn, so we know that 3 and 5 a2 a3 a5 can each divide n at most once. Thus we have n = 2 3 5 , and a3; a5 are either 0 or 1. a2 a2 a2−1 Suppose a3 = a5 = 0 so that n = 2 . Then φ(n) = φ(2 ) = 2 (2 − 1), which impiles that a2 = 4; n = 16. a2 a2 a2−1 0 Suppose a3 = 1; a5 = 0, so that n = 2 ·3. Then φ(n) = φ(2 ·3) = 2 (2−1)3 (3−1) = a2 2 . This implies that a2 = 3; n = 8 · 3 = 24. a2 a2 a2−1 0 Suppose a3 = 0; a5 = 1, so that n = 2 ·5. Then φ(n) = φ(2 ·5) = 2 (2−1)5 (5−1) = a2+1 2 . This implies that a2 = 2; n = 4 · 5 = 20. a2 a2 Suppose a3 = 1; a5 = 1, so that n = 2 · 3 · 5. If a2 > 0 then φ(n) = φ(2 · 3 · 5) = a2−1 0 0 a2+2 2 (2 − 1)3 (3 − 1)5 (5 − 1) = 2 . This implies that a2 = 1; n = 2 · 3 · 5 = 30. If a2 = 0 then instead φ(n) = φ(3 · 5) = 2 · 4 = 8 does in fact work, so n = 15. Thus the possibilities are n = 15; 16; 20; 24; 30. 5.2 Summatory functions For further reading on the material in this subsection, consult Rosen 7.2, Shoup 2.9. In this section we'll discuss another class of multiplicative functions, known as summatory functions. Though these do not look like they should be multiplicative, they often are. Definition 5.11. If f is an arithmetic function, we define the summatory function of f to be X F (n) = f(d) djn http://jaydaigle.net/teaching/courses/2016-fall-322/ 60 Jay Daigle Occidental College Math 322: Number Theory where the sum is over all numbers d which divide n. Definition 5.12. We define the number of divisors function τ(n) to be the number of natural P numbers ≤ n which divide n.