The Euler Phi Function

The Euler Phi Function 7-3-2006 An arithmetic function takes ositive integers as inuts and roduces real or comlex numbers as oututs. If f is an arithmetic function, the divisor sum Dfn) is the sum of the values of f at the ositive divisors of n. τn) is the number of ositive divisors of n; σn) is the sum of the ositive divisors of n. The Möbius function µn) is if n = and 0 if n has a reeated rime factor. Otherwise, it is ) k, where k is the number of distinct) rime factors. The Dirichlet roduct of arithmetic functions f and g is f g)n) = n fd)g. d) The Möbius inversion formula says that µ Df = f. Dφn) = n. φn) = n. The Euler hi function is multilicative: If m, n) =, then φmn) = φm)φn). I m going to describe a formula for comuting φn) which uses the rime factorization of n. This will require some reliminaries on divisor sums. Definition. An arithmetic function is a function defined on the ositive integers which takes values in the real or comlex numbers. Examles. a) Define f : Z + R by fn) = sin n. Then f is an arithmetic function. b) The Euler hi function φ is an arithmetic function. c) Define τ : Z + Z + by τn) = the number of ositive divisors of n). For examle, τ2) = 6, since there are 6 ositive divisors of 2, 2, 3, 4, 6, and 2. τ is an arithmetic function. d) Define σ : Z + Z + by σn) = the sum of the ositive divisors of n). Since, 2, 3, 6, 9, and 8 are the ositive divisors of 8, σ is an arithmetic function. σ8) = + 2 + 3 + 6 + 9 + 8 = 39.

Definition. The Möbius function is the arithmetic function defined by µ) =, and for n >, { µn) = ) k if n = i k, i distinct rimes 0 otherwise. Thus, µn) = 0 if n is divisible by a square. Examle. µ6) =, since 6 = 2 3. Likewise, µ30) =, since 30 = 2 3 5. But µ2) = 0 and µ250 = 0. Definition. If f is an arithmetic function, the divisor sum of f is [Df)]n) = fd). To save writing, I ll make the convention that when I write, I mean to sum over all the ositive divisors of a ositive integer n. Thus, the divisor sum of f evaluated at a ositive integer n takes the ositive divisors of n, lugs them into f, and adds u the results. Notice that the divisor sum is a function which takes an arithmetic function as inut and roduces an arithmetic function as outut. The divisor sum machine takes in an arithmetic function f. With f installed in the divisor sum machine, you get a new arithmetic function: Df). f arithmetic function Divisor Sum Divisor Sum f Df) works by taking a number, alying f to each divisor of the sum, and adding u the results. 6 Divisor Sum f 2 3 6 f) + f2) + f3) + f6) 2

Examle. Suose f : Z + Z + is defined by fn) = n 2. Then [Df)]n) = d 2. For examle, [Df)]2) = d 2d 2 = 2 + 2 2 + 3 2 + 4 2 + 6 2 + 2 2 = 20. Lemma. [Dµ)]n) = µd) = { if n = 0 otherwise Proof. The formula for n = is obvious. Suose n >. Let n = r rk k be the factorization of n into distinct rimes. What are the nonzero terms in the sum µd)? They will come from d s which are roducts of single owers of,... k, and also d =. For examle, µ 2 7 ) and µ 2 4 ) would give rise to nonzero terms in the sum, but µ 3 3 8 ) = 0. So µd) = + µ ) + + µ k )) + µ 2 ) + µ 3 ) + + µ k k )) + + µ 2 k ) = + ) k ) + ) k ) 2 + + 2 ) k ) k = ) k = 0. k Examle. Suose n = 24. The divisor sum is µd) = µ) + µ2) + µ3) + µ4) + µ6) + µ2) + µ24) = + ) + ) + 0 + + 0 + 0 = 0. d 24 Definition. If f and g are arithmetic functions, their Dirichlet roduct is f g)n) = n fd)g. d) Examle. If f and g are arithmetic functions, f g)2) = f)g2) + f2)g6) + f3)g4) + f4)g3) + f6)g2) + f2)g). I ll need two heler functions in what follows. Define In) = for all n Z +, 3

{ if n = en) = 0 otherwise Proerties of the Dirichlet roduct. for all n Z +. Let f, g, and h be arithmetic functions.. f g = g f. 2. f g) h = f g h). 3. f e = f = e f. 4. f I = Df = I f. 5. µ I = e. Proof. For roerty, note that divisors of n come in airs { n } is d, d. This means that the same terms occur in both f g)n) = n fd)g d) { d, n } {, and that if d, n } is a divisor air, so d d and g f)n) = gd)f n d), so they re equal. Associativity is a little tedious, so I ll just note that [f g) h]n) and [f g h)]n) are equal to {d,e,f} fd)ge)hf), where the sum runs over all triles of ositive numbers d, e, f such that def = n. You can fill in the details. For roerty 3, note that f e)n) = n fd)e = fn)e) = fn). d) n e is 0 excet when d) n =, i.e. when d = n.) d For roerty 4, f I)n) = n fd)i = d) fd) = fd) = Df)n). For roerty 5, start with n = : µ I)) = µ)i) = = = e). Now suose n > 0. Then µ I)n) = Dµ)n) = 0 = en). Therefore, the formula holds for all n. The next result is very owerful, but the roof will look easy with all the machinery I ve collected. Theorem. Möbius Inversion Formula) If f is an arithmetic function, then f = µ Df. Proof. µ Df = µ I f = e f = f. Next, I ll comute the divisor sum of the Euler hi function. 4

Lemma. [Dφ)]n) = φd) = n. Proof. Let n be a ositive integer. Construct the fractions n, 2 n,..., n n, n n. Reduce them all to lowest terms. Consider a tyical lowest-term fraction a. Here d n because it d came from a fraction whose denominator was n, a < d because the original fraction was less than ), and a, d) = because it s in lowest terms). Notice that going the other way) if a is a fraction with ositive to and bottom which satisfies d n, d a < d, and a, d) =, then it is one of the lowest-terms fractions. For dk = n for some k, and then a d = ka kd = ka and the last fraction is one of the original fractions. n How many of the lowest-terms fractions have d on the bottom? Since the a on to is a ositive number relatively rime to d, there are φd) such fractions. Summing over all d s which divide n gives φd). But since every lowest-terms fraction has some such d on the bottom, this sum accounts for all the fractions and there are n of them. Therefore, φd) = n. Examle. Suose n = 6. Then φd) = φ) + φ2) + φ3) + φ6) = + + 2 + 2 = 6. d 6 Lemma. Let n. φn) = µd) n d. Proof. By Möbius inversion and the revious result, φn) = µ Dφ)n) = n µd)dφ = d) µd) n d. Examle. Take n = 6, so φ6) = 2. d 6 µd) 6 d = µ) 6 + µ2) 6 2 + µ3) 6 3 + µ6) 6 6 = )6) + )3) + )2) + )) = 2. Theorem. Let n. φn) = n 5

By convention, the emty roduct the roduct with no terms equals.) Proof. If n =, the result is immediate by convention. If n >, let,..., k be the distinct rime factors of n. Then ) ) = ) 2 k = i i + i j + ) k. i j 2 k Each term is ± d, where d is the first term) or a roduct of distinct rimes. The )i in front of each term alternates signs according to the number of s which is exactly what the Möbius function does. So the exression above is µd) d. I can run the sum over all divisors, because µd) = 0 if d has a reeated rime factor.) Now simly multily by n: n = µd) n d = φn). Examle. 40 = 2 3 5, so Likewise, 8 = 3 4, so φ40) = 40 = 6. 2 5 φ8) = 8 = 54. 3 More generally, if is rime and k, then φ k ) = k k. For φ k ) = k = k k. Definition. An arithmetic function f is multilicative if m, n) = imlies fmn) = fm)fn). Lemma. φ is multilicative that is, if m, n) =, then φmn) = φm)φn). Proof. Suose m, n) =. Now φm) = m m and φn) = n q n q rime. q 6

So φm)φn) mn = m ) q n q rime q ). Since m, n) =, the two roducts have no rimes in common. Moreover, the rimes that aear in either of the roducts are exactly the rime factors of mn. So φm)φn) mn = r mn r rime. r Hence, φm)φn) = mn) r mn r rime = φmn). r Examle. If n 3, then φn) is even. In fact, if n has k odd rime factors, then 2 k φn). To see this, observe first that φ2 k ) = 2 k 2 k. This is even if 2 k 4. So suose that n has k odd rime factors. Then φn) = n = φn) = n ) = n ). The denominator of the fraction is the roduct of rimes dividing n, so the fraction is actually an integer. The second term has at least one even factor for each odd rime dividing n, because if is an odd rime then is even. Hence, the second term and therefore φn) is divisible by 2 k. For examle, consider 7623 = 3 2 7 2. There are 3 odd rime factors, so φ7623) should be divisible by 8. And in fact, φ7623) = 3960 = 8 495. c 2006 by Bruce Ikenaga 7

The Sum and Number of Divisors 7-3-2006 Definition. The sum of divisors function is given by σn) = d. The number of divisors function is given by τn) =. Examle. Recall that a number is erfect if it s equal to the sum of its divisors other than itself. It follows that a number n is erfect if σn) = 2n. Examle. σ5) = + 3 + 5 + 5 = 24 and τ5) = 4. I want to show that σ and τ are multilicative. I can do most of the work in the following lemma. Lemma. The divisor sum of a multilicative function is multilicative. Proof. Suose f is multilicative, and let Df) be the divisor sum of f. Suose m, n) =. Then [Df)]m) = a m fa) and [Df)]n) = b n fb). Then [Df)]m) [Df)]n) = fa) fb) = fa)fb). a m b n a m b n Now m, n) =, so if a m and b n, then a, b) =. Therefore, multilicativity of f imlies [Df)]m) [Df)]n) = fab). a m b n Now every divisor d of mn can be written as d = ab, where a m and b n. Going the other way, if a m and b n then ab mn. So I may set d = ab, where d mn, and relace the double sum with a single sum: [Df)]m) [Df)]n) = d mn fab) = [Df)]mn). This roves that Df) is multilicative. Examle. The identity function idx) = x is multilicative: idmn) = mn = idm) idn) for all m, n so a fortiori for m, n) = ). Therefore, the divisor sum of id is multilicative. But [Did)]n) = idd) = d = σn).

Hence, the sum of divisors function σ is multilicative. Examle. The constant function In) = is multilicative: Imn) = mn = Im) In) for all m, n so a fortiori for m, n) = ). Therefore, the divisor sum of I is multilicative. But [DI)]n) = Id) = = τn). Hence, the number of divisors function τ is multilicative. I ll use multilicativity to obtain formulas for σn) and τn) in terms of their rime factorizations as I did with φ). First, I ll get the formulas in the case where n is a ower of a rime. Lemma. Let be rime. Then: σ k ) = k+ τ k ) = k + Proof. The divisors of k are,, 2,..., k. So the sum of the divisors is σ k ) = + + 2 + + k = k+. And since the divisors of k are,, 2,..., k, there are k + of them, and τ k ) = k +. Theorem. Let n = r rk k, where the s are distinct rimes and r i for all i. Then: r + σn) = ) ) k k rk+ τn) = r + ) r k + ) Proof. These results follow from the receding lemma, the fact that σ and τ are multilicative, and the fact that the rime ower factors ri i are airwise relatively rime. Here is a grah of σn) for n 000. 3000 2500 2000 500 000 500 200 400 600 800 000 2

Note that if is rime, σ = +. This gives the oint, +), which lies on the line y = x +. This is the line that you see bounding the dots below. Here is a grah of τn) for n 000. 30 25 20 5 0 5 200 400 600 800 000 If is rime, τ) = 2. Thus, τ reeatedly returns to the horizontal line y = 2, which you can see bounding the dots below. Examle. 720 = 2 4 3 2 5, so σ720) = 2 5 2 ) 3 3 3 ) 5 2 ) = 248, 5 τ720) = 4 + )2 + ) + ) = 30. Examle. For each n, there are only finitely many numbers k whose divisors sum to n: σk) = n. For k divides itself, so n = σk) = junk) + k > k. This says that k must be less than n. So if I m looking for numbers whose divisors sum to n, I only need to look at numbers less than n. For examle, if I want to find all numbers whose divisors sum to 42, I only need to look at {, 2,..., 4}. c 2006 by Bruce Ikenaga 3

Perfect Numbers 7-3-2006 Definition. A number n > 0 is erfect if σn) = 2n. Equivalently, n is erfect if it is equal to the sum of its divisors other than itself. Examle. 6 is erfect, because 6 = + 2 + 3, or 2 6 = + 2 + 3 + 6. It is not known whether there are any odd erfect numbers, or whether there are infinitely many even erfect numbers. The existence of infinitely many even erfect numbers is related to the existence of infinitely many Mersenne rimes by the following result. Proosition. n is an even erfect number if and only if n = 2 k 2 k ), where 2 k is a Mersenne rime. Proof. First, suose 2 k is rime. Then n = 2 k 2 k ) is even; I want to show that it s erfect. Since 2 k is an odd rime, it is relatively rime to 2 k. Hence, σn) = σ 2 k 2 k ) ) = σ 2 k ) σ 2 k ) 2 k ) 2 k ) 2 ) = 2 2 k = ) 2 k ) 2 k ) + ) = 2 k )2 k = 2 2 k 2 k ) = 2n. Therefore, n is erfect. Conversely, suose n is an even erfect number. I want to show n = 2 k 2 k ), where 2 k is a Mersenne rime. Since n is even, I can write n = 2 i m, where i and m is odd. Then 2 i+ m = 2n = σn) = σ2 i m) = σ2 i )σm) = 2 i+ )σm). Since 2 i+ divides the left side, it divides the right side. But 2 i+ is odd, so I must have 2 i+ σm). I claim further that 2 i+ is the highest ower of 2 which divides σm). For if 2 i+2 σm), then 2 i+ m = 2 i+ )σm) = 2 i+ ) 2 i+2 junk. Hence, m = 2 i+ ) 2 junk, which contradicts the fact that m is odd. Since I now know that 2 i+ is the highest ower of 2 which divides σm), I can write σm) = 2 i+ s, where s is odd. Then Hence, 2 i+ m = 2 i+ )σm) = 2 i+ ) 2 i+ s, so m = 2 i+ )s. n = 2 i m = 2 i 2 i+ )s. If I can show s =, then I will have gotten n to have the right form. To do this, start with m = 2 i+ )s. Add s to both sides to get m + s = 2 i+ s = σm). m is divisible by, by itself, and by s because m = 2 i+ )s). If s = m, then n = 2 i m = 2 i 2 i+ )s = 2 i 2 i+ )m, so = 2 i+.

so This imlies i = 0 տր. So s m. If in addition s >, then, s, and m are three distinct divisors of m, σm) m + s +. This contradicts m + s = σm), derived above. Therefore, s =. At this oint, I know n = 2 i 2 i+ ). I only need to show that 2 i+ is rime. Since and 2 i+ are distinct factors of 2 i+, I have 2 i+ = σm) = σ2 i+ ) + 2 i+ ) = 2 i+. Therefore, σ2 i+ ) = 2 i+. But this means that and 2 i+ are the only factors of 2 i+, i.e. 2 i+ is rime. Examle. 2 7 = 27 is rime, so is erfect. 2 6 2 7 ) = 828 I now know that finding even erfect numbers is equivalent to finding Mersenne rimes rimes of the form 2 n. I showed earlier that 2 n is rime imlies that n is rime. So to look for Mersenne rimes, I only need to look at 2 n for n rime. Next, I ll derive a result which simlifies checking that 2 n is rime. First, here s an amusing lemma. Lemma. 2 a, 2 b ) = 2 a,b). Proof. Assume without loss of generality that a b. The greatest common divisor of two numbers doesn t change if I subtract the smaller from the larger, so 2 a, 2 b ) = 2 a ) 2 b ), 2 b ) = 2 a 2 b, 2 b ) = 2 b 2 a b ), 2 b ). Since 2 b is odd, it has no factors in common with the 2 b in the first term. So 2 b 2 a b ), 2 b ) = 2 a b, 2 b ). Now I see that the 2 ) in each slot is just along for the ride: All the action is taking lace in the exonents. And what is haening is that the subtraction algorithm for comuting greatest common divisors is oerating in the exonents! the original air a, b, was relaced by a b, b. It follows that the exonents will converge to a, b), because this is what the subtraction algorithm does. And when the algorithm terminates, I ll have 2 a,b), 0) = 2 a,b), roving the result. Examle. 42, 54) = 6, so 2 42, 2 54 ) = 2 6 = 63. This is surely not obvious, esecially when you consider that 2 42 = 4398046503 and 2 54 = 80439850948983! Theorem. Let be an odd rime. Every factor of 2 has the form 2k + for some k 0. Proof. It suffices to rove that the result holds for rime factors of 2. For 2a + )2b + ) = 22ab + a + b) +, 2

so roducts of numbers of the form 2k + also have that form. Suose then that q is a rime factor of 2. Little Fermat says q 2 q. The receding lemma imlies that 2, 2 q ) = 2,q ). Now q 2 and q 2 q imlies q 2,q ). In articular, 2,q ) >, since it s divisible by the rime q. This in turn imlies that, q ) >. Now is rime, so this is only ossible if, q ) =. In articular, q. Write q = t, so q = t +. q is odd, so q is even, and t is even. Since is odd, t must be even: t = 2k for some k. Then q = 2k +, which is what I wanted to show. Examle. Is 2 7 = 307 rime? 307 362. If 2 7 has a roer rime factor, it must have one less than 362, and the rime factor must have the form 2k 7 + = 34k +. So I need to check the rimes less than 362 to see if they divide 307. k 34k + 35 Not rime 2 69 Not rime 3 03 03 307 4 37 37 307 5 7 Not rime 6 205 Not rime 7 239 239 307 8 273 Not rime 9 307 307 307 0 34 Not rime Therefore, 2 7 is rime. c 2006 by Bruce Ikenaga 3