HOW OFTEN IS EULER S TOTIENT A PERFECT POWER? 1. Introduction

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1 HOW OFTEN IS EULER S TOTIENT A PERFECT POWER? PAUL POLLACK Abstract. Fix an integer k 2. We investigate the number of n x for which ϕn) is a perfect kth power. If we assume plausible conjectures on the distribution of smooth shifted primes, then the count of such n is at least x/lx) +o), as x, where Lx) = exp log log / log ). This lower bound is implicit in work of Banks Friedlander Pomerance Shparlinski. We prove unconditionally that x/lx) +o) serves as an upper bound. In fact, we establish this same bound for the count of n x for which ϕn) is squarefull. The proof builds on methods recently introduced by the author to study popular subsets for Euler s function.. Introduction Let ϕn) denote Euler s totient function, i.e., ϕn) = #Z/nZ). One of several problems studied in the paper [2] of Banks, Friedlander, Pomerance, and Shparlinski is the question of how often ϕn) assumes square values. The methods of [2] for obtaining lower bounds in this problem depend on the distribution of smooth shifted primes. Recall that a positive integer n is called Y -smooth alternatively, Y -friable) if there are no primes dividing n that exceed Y. We write ΨX, Y ) for the count of Y -smooth n X. It is natural to guess that shifted primes p are smooth with roughly the same frequency as ordinary integers of the same size. This thought, which in vague form goes back at least to Erdős s work in [4], led Pomerance to float the following precise conjecture in [0]. Conjecture. If X, Y with X Y, then #{p X : p is Y -smooth} ΨX, Y ) log X. Conjecture appears rather difficult, and may even be overly optimistic. But the authors of [2] show that even much weaker statements in this direction have interesting consequences for the distribution of square values of ϕn). Specifically, suppose that for a certain fixed U > and a certain real number K, we have that ) #{p X : p is X /U -smooth} ΨX, X/U ) log X) K for all large X. Then it is shown in [2] that there are at least x /U+o) values of n x for which ϕn) is a square, as x. Baker and Harman have proved that ) holds for any U Consequently, ϕn) is a square for at least x values of n x. Conjecture would predict that any U > is admissible in ), so that there are more than x ɛ values of n x with ϕn) a square. 200 Mathematics Subject Classification. N64 primary), N25, N37 secondary).

2 2 PAUL POLLACK Running these arguments with U allowed to vary, we can obtain a sharper lower bound than x ɛ. This is the substance of our first theorem, which also treats kth powers rather than merely squares. Writing log j for the jth iterate of the natural logarithm, let ) log3 x Lx) = exp. log 2 x Theorem conditional on Conjecture ). Fix an integer k 2. As x, the number of n x for which ϕn) is a kth power is at least x/lx) +o). It will emerge from the proof that the full strength of Conjecture is not needed; all we need is a lower bound of roughly the predicted size when Y exp log X).) While Theorem does not appear explicitly in [2], our argument below is a straightforward modification of the proof of Theorem 6.4 in [2]. We include the proof for completeness, and for comparison with our main result, which we now turn to. The discussion so far has focused on lower bounds. As regards upper bounds, it is proved in [2] that as x ) the number of n x for which ϕn) is squarefull does not exceed x exp + o)) log log ). Of course, this serves as an upper bound on the number of n x for which ϕn) is a kth power for any k 2). There is quite a gap between this upper bound and the lower bound of x/lx) +o) in Theorem. The main purpose of this note is to close this gap. Theorem 2. As x, the number of n x for which ϕn) is squarefull is at most x/lx) +o). The proof of Theorem 2 uses methods recently introduced by the author in [8] to study popular values of Euler s function cf. [0,, 2]). The cognate problem of counting squarefull m in the range of the Euler function rather than their preimages, as done here) was recently studied in [9]. It was shown there that x/) 3 #{squarefull m x 2 : m ϕn)} x/) In fact, the lower bound holds for the number of squares up to x 2 in the range of ϕ. For related results, see [3, 5, 6]. We note that the expression x/lx) +o) also arises in the study of counterexamples to the converse of Fermat s little theorem. Conjecturally, it describes the count of Carmichael numbers up to x, as well the count of base a pseudoprimes for each fixed a 0, ±. For Carmichael numbers, the upper bound implicit in the conjecture is known, while for base a pseudoprimes, an upper bound of x/lx) /2 for x > x 0 a)) has been shown. The heuristic lower bound arguments invoke a variant of Conjecture. For details, see []. Notation. The letter p is reserved throughout for primes. For Y > 0, we define arithmetic functions ω >Y ) and Ω >Y ) by ω >Y n) = p n p>y, and Ω >Y n) = p e n p>y e;

3 HOW OFTEN IS EULER S TOTIENT A PERFECT POWER? 3 when Y = 0, we omit the subscripts and write simply ω ) and Ω ). 2. Proof of Theorem The following result in combinatorial group theory follows from combining Theorem. and Proposition.2 of []. Lemma 3. Let G be a finite abelian group of exponent m, and let d = m + log G /m)). For each pair of integers r, t) with r > t > d, any sequence of r elements of G contains at least r t) / r d) distinct subsequences of length at most t and at least t d, whose product is the identity element. Proof of Theorem. Let X = explog 2 x) 2 ) and y = /log 2 x) 3. Let r denote the number of primes p X for which p is y-smooth. We will construct the values of n in Theorem as certain squarefree products of at most t of these r primes, where t := /log 2 x) 2. Note that all such products are at most x. For each of the r primes p described above, write p = l y lv lp ), where l runs over the primes at most y. We associate to p the mod k)-reduced exponent vector v l p ) mod k) l y Z/kZ) πy). Asking for the product n of t of these primes to have ϕn) a kth power amounts to asking that the sum of the corresponding exponent vectors be the zero element in G := Z/kZ) πy). This implies that the number of n as in Theorem is at least r ) t / r ) d, where d = dg) is the quantity of Lemma 3. Assuming Conjecture, we have for large x that r ΨX, y)/ log X. Now applying standard estimates for Ψ, ) for example, Theorem 2. of [2]), we find that, with U := log X log y, r Since G has exponent k, X exp + o))u log U) log X = X exp + o)) log 2 x log 3 x). d + k + logk πy) )) < y once x is large enough). Thus, r d) r d X y = Lx) o). On the other hand, ) r r ) t = r t t t r t Lx) o) X t /Lx) +o) = x/lx) +o). t t It follows that r t) / r d) x/lx) +o), as desired.

4 4 PAUL POLLACK Remark. The kth powers appearing above are all y-smooth. There are only Lx) o) y-smooth numbers n x see, e.g., Theorem 5.2 of [3]), so by the pigeonhole principle, some kth power m x satisfies #ϕ m) x/lx) +o). Compare with the first half of [2, Theorem 6.2].) Pomerance [0] has shown that max m x #ϕ m) x/lx) +o). We conclude that there are kth powers that are essentially as popular as possible as ϕ-values. 3. Preliminaries for the proof of Theorem 2 We quote several results from the author s recent work [8]. Given a real number z 2, we let W z ) denote the additive arithmetic function whose value at each prime power p e is given by W z p e ) = ω >z ϕp e )). Lemma 4 [8, Lemma 2]). Let Fix any η 0, ), and let As x, z = explog 2 x) /2 ). A = log 2 x) η. A Wzn) xlx) o). n x Lemma 5 [8, Lemma 3]). Let Y, Z. The number of positive integers n x with Ω >Y p ) ω >Y p )) Z p n is at most xlx) 2+o) Y Z/2, as x uniformly in Y, Z). Lemma 6 [8, Lemma 4]). Let Z. The number of n x with 2) ωn) log 3 x log 2 x) 2 satisfying is at most as x uniformly in Z). Ωp ) Z p n xlx) 2+o) 2 Z/2, Lemma 7 see [8, Lemma 5] and the remark following). The following statement holds for a certain constant C > 0: Let x 3, and let d be a positive integer. Then the number of positive integers n x for which d ϕn) is at most x d Clog 2 x) 2 ) Ωd) Ωd) Ωd). We also require the following exponential moment estimate, which will be used to bound the number of integers possessing many prime factors exceeding.

5 Lemma 8. Let y := uniformly for real T [, x]. HOW OFTEN IS EULER S TOTIENT A PERFECT POWER? 5. Then, as x, log 2 x) 2 y Ω >n) T x o), n T Proof. Let G be the multiplicative function determined by the convolution identity y Ω >n) = d n Gd). Then G vanishes at prime powers p a with p, while if p >, then Gp a ) = y a y a. Hence, y Ω >n) = T Gd) d n T d T T Gd) T + y ) + y2 y d p p 2 d T <p T If T, the product is empty and thus equal to ). So suppose that T >. For all p, T ], we have that assuming x is large enough) and so <p T y p + y2 y p 2 + = y p y < y p/2 < 2y p, + y ) + y2 y ) ) y +... exp + y2 y +... p p 2 p p 2 <p T ) 2y exp < exp 2y log p 2 T ) exp2 / log 2 x) = x o). <p T Collecting the estimates completes the proof. 4. Proof of Theorem 2 In this section, z has the same meaning as in Lemma 4: z = explog 2 x) /2 ). We let ɛ > 0 be arbitrary but fixed. We begin by excluding a set of inconvenient values of n, of cardinality Ox/Lx) 2ɛ ). This part of the argument closely parallels work in [8], but we repeat the details for the convenience of the reader. We start by assuming that ωn) satisfies the inequality 2). By a 97 theorem of Hardy Ramanujan [7], the number of n x with ωn) > log 3 x/log 2 x) 2 is x log 2 x + O))k x/lx) +o), k )! k> log 3 x/log 2 x) 2 as x, which is well within our target upper bound. Next, we assume that ϕn) > x/lx). Since ϕn) n/ log 2 n, the number of n x with ϕn) x/lx) is Ox log 2 x/lx)), which is acceptable.

6 6 PAUL POLLACK We further assume that W z n) ɛ) log 2 x. By Lemma 4, with η = ɛ, this excludes at most x/lx) ɛ)2 +o) values of x as x ). Again, this is acceptable. Next, we assume that 3) Ω z ϕn)) If this inequality fails, then 2 ɛ log 2 x < Ω >zϕn)) W z n) = p e nω >z ϕp e )) ω >z ϕp e ))) 2 ɛ ) log 2 x. = Ω >z p e ) ω >z p e )) + p e n p n Ω >z p ) ω >z p )) = e 2) + p e n p n e 2 p>z Ω >z p ) ω >z p )). Thus, one of the final two sums exceeds ɛ 4 log 2 x p e n e 2 p e p e n e 2 p>z z e > z 4 ɛ log 2 x = exp. If it is the first sum, then ) 4 ɛ. log 2 x) /2 The number of n x with a squarefull divisor of this size does not exceed x m x exp ) 8 ɛ. log 2 x) /2 ) m>exp 4 ɛ log 2 x) /2 m squarefull The sum on m has been estimated by partial summation, after recalling that the count of squarefull m T is OT /2 ), for all T.) In particular, the number of these n is smaller than x/lx) for large x. If the second of the two sums exceeds ɛ, then Lemma 5 with Y = z and Z = ɛ ) shows that n belongs to a set 4 log 2 x 4 log 2 x of size at most xlx) 2+o) Y Z/2 = xlx) 2+o) exp 8 ɛ log 2 x) /2 which is again smaller than x/lx) for large x. Finally, we assume that 4) Ωϕn)) log 2 x) 2/3. Supposing this inequality fails, we have that Ωp ) + Ωϕp e )) = Ωϕn)) > p n p e n, e 2e ) = p n ), log 2 x) 2/3.

7 HOW OFTEN IS EULER S TOTIENT A PERFECT POWER? 7 So one of the two sums on the far left exceeds. If it is the latter, then n 2 log 2 x) 2/3 is divisible by a squarefull number exceeding 2 2 /log 2 x)2/3. The number of such n x does not exceed x m x exp log 2 ) ; 4 log 2 x) 2/3 m>2 2 /log 2 x)2/3 m squarefull this is smaller than x/lx) for large x. If it is the former, we apply Lemma 6 to see that n is placed in a certain set of size at most xlx) 2+o) exp log 2 ), 4 log 2 x) 2/3 which is again smaller than x/lx) for large x. We will show that the number of n x for which ϕn) is squarefull, and for which all of the above assumptions are satisfied, is smaller than x 0 ɛ. As this is ox/lx)), we conclude that the total number of n x with ϕn) squarefull is Ox/Lx) 2ɛ ). Since ɛ may be taken arbitrarily small, the theorem follows. Every squarefull integer can be decomposed as a 2 b 3, with a, b positive integers. Hence, we may write where ϕn) = a 2 a 2 2a 2 3b 3 b 3 2b 3 3 p a b p z, p a 2 b 2 z < p, p a 3 b 3 < p. From the definition of z together with 4), ) a 2 b 3 z Ωa2 b3 ) z Ωϕn)) exp = x o). log 2 x) /6 Since ϕn) > x/lx), we deduce that 5) a 2 2a 2 3b 3 2b 3 3 > x Lx)a 2 b 3 = x +o). Put R = and δ = /R. Choose α 2, α 3, β 2, β 3 {/R, 2/R,..., R )/R, } such that, for i = 2, 3, a i [x α i δ, x α i ], while b i [x β i δ, x β i ]. Note that x /R, and so a i x α i, b i x β i. By 5), 6) 2α 2 + 2α 3 + 3β 2 + 3β 3 + o). Also, recalling 3), 7) 2Ωa 2 ) + 2Ωa 3 ) + 3Ωb 2 ) + 3Ωb 3 ) = Ω z ϕn)) 2 ɛ ) log 2 x.

8 8 PAUL POLLACK Switching perspective, we may view a, a 2, a 3, b, b 2, b 3 as given, and count the number of corresponding values of n x. By Lemma 7, #{n x : ϕn) = a 2 a 2 2a 2 3b 3 b 3 2b 3 3} #{n x : a 2 3b 3 3 ϕn)} x Clog a 2 3b 3 2 x) 2 ) Ωa2 3 b3 3 ) Ωa 2 3b 3 3) Ωa2 3 b3 3 ). 3 Keeping 7) in mind, and using that a i x α i, b i x β i, this upper bound is seen to be at most x 2α 3 3β 3 +o) ) Ωa2 3 b3 3 ) = x 2α 3 3β 3 +o) ) 2A 3+3B 3, where we define A i = Ωa i ) and B i = Ωb i ). Our strategy for the rest of the proof is as follows: We fix values for α 3, β 3, A 3, and B 3. This puts an additional constraint on the tuples a, a 2, a 3, b, b 2, b 3 ; namely, it amounts to specifying the order of magnitude and the number of prime factors of a 3 and b 3. In the next paragraph, we bound from above the number of tuples a, a 2, a 3, b, b 2, b 3 obeying this constraint. Each such tuple corresponds to at most x 2α 3 3β 3 +o) ) 2A 3+B 3 values of n. Thus, multiplying our bound on the number of these tuples by x 2α 3 3β 3 +o) ) 2A 3+B 3 yields a bound on the number of n corresponding to this particular choice of α 3, β 3, A 3, and B 3. As we will see, some additional effort is required to suss out the size of this bound.) Finally, we bound the total number of n by summing on α 3, β 3, A 3, and B 3. We emphasize that all o) terms appearing here and below are to be understood with x, and that the convergence of o) to 0 is uniform in all other parameters. Let y be as in Lemma 8, i.e., y =. The total number of )-smooth log 2 x) 2 integers in [, x] is x o) see again Theorem 5.2 of [3]), and so there are at most x o) possibilities for each of a, b, a 2, b 2. We have that a 3 x α 3 with α 3 fixed), that every prime factor of a 3 exceeds, and that Ωa 3 ) takes the fixed value A 3 ; thus, the number of possibilities for a 3 is at most y A 3 y Ω >m) y A 3 x α3+o). m x α 3 Similarly, the number of possibilities for b 3 is at most y B 3 x β 3+o). Thus, the number of n corresponding to a fixed choice of α 3, β 3, A 3, B 3 is at most 8) x 2α 3 3β 3 +o) ) 2A 3+3B3 x α 3+β 3 +o) y A 3 B 3. Assume that there is some tuple a, a 2, a 3, b, b 2, b 3 corresponding to our fixed choice of α 3, β 3, A 3, and B 3. If this fails, then there are no corresponding n.) Choose one and define, for i = 2, 3, α i = A i / log 2 x, and βi = B i / log 2 x. Since A 3, B 3 are fixed, the values of α 3 and β 3 do not depend on the choice of tuple a, a 2, a 3, b, b 2, b 3, but the values of α 2 and β 2 may.) By 7), 9) 2 α α β β 3 2 ɛ. For both i = 2, 3, we have that ) A i = x α i, ) B i = x β i, y A i = x α i+o), and y B i = x β i +o) as x ); making these substitutions above transforms our upper

9 bound 8) into HOW OFTEN IS EULER S TOTIENT A PERFECT POWER? 9 0) x α 3 α 3 ) 2β 3 β 3 )+o). How large is the exponent here? Subtracting 9) from 6), hence, 2α 2 α 2 ) + 2α 3 α 3 ) + 3β 2 β 2 ) + 3β 3 β 3 ) 2 ɛ + o); 2α 3 α 3 ) + 3β 3 β 3 ) 2 ɛ + 2 α 2 α 2 ) + 3 β 2 β 2 ) + o). Since every prime factor of a 2 b 2 is at most, we see that x α 2 δ a 2 ) A 2 = x α 2 ; thus, α 2 α 2 + o). Similarly, β 2 β 2 + o). Therefore, 2α 3 α 3 ) + 3β 3 β 3 ) 2 ɛ + o). Since each prime factor of a 3, b 3 exceeds, we find that α 3 α 3 and β 3 β 3 ; so both of the left-hand summands in the last display are nonnegative. Thus, either α 3 α 3 ɛ + o) or β 8 3 β 3 ɛ + o). Putting this back into 0), we find that 2 the number of n corresponding to a fixed choice of α 3, β 3, A 3, B 3 is at most x 8 ɛ+o). It remains to sum on α 3, β 3, A 3, B 3. There are crudely) only O) possibilities for each of these parameters. Hence, the total number of n satisfying our assumptions for which ϕn) squarefull is at most x 8 ɛ+o) ) 4, and so is smaller than x 0 ɛ for large x, as desired. Acknowledgements The author thanks Carl Pomerance and the anonymous referee for helpful comments. Work of the author is supported by NSF award DMS References. W. R. Alford, A. Granville, and C. Pomerance, There are infinitely many Carmichael numbers, Ann. of Math. 2) ), W. D. Banks, J. B. Friedlander, C. Pomerance, and I. E. Shparlinski, Multiplicative structure of values of the Euler function, High primes and misdemeanours: lectures in honour of the 60th birthday of Hugh Cowie Williams, Fields Inst. Commun., vol. 4, Amer. Math. Soc., Providence, RI, 2004, pp W. D. Banks and F. Luca, Power totients with almost primes, Integers 20), P. Erdős, On the normal number of prime factors of p and some related problems concerning Euler s ϕ-function, Quart. J. Math ), T. Freiberg, Products of shifted primes simultaneously taking perfect power values, J. Aust. Math. Soc ), T. Freiberg and C. Pomerance, A note on square totients, Int. J. Number Theory 205), G. H. Hardy and S. Ramanujan, The normal number of prime factors of a number n, Quart. J. Math ), P. Pollack, Popular subsets for Euler s ϕ-function, submitted. Preprint available from http: //pollack.uga.edu/research.html. 9. P. Pollack and C. Pomerance, Square values of Euler s function, Bull. Lond. Math. Soc ),

10 0 PAUL POLLACK 0. C. Pomerance, Popular values of Euler s function, Mathematika ), , On the distribution of pseudoprimes, Math. Comp ), , Two methods in elementary analytic number theory, Number theory and applications Banff, AB, 988), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., vol. 265, Kluwer Acad. Publ., Dordrecht, 989, pp G. Tenenbaum, Introduction to analytic and probabilistic number theory, third ed., Graduate Studies in Mathematics, vol. 63, American Mathematical Society, Providence, RI, 205. Department of Mathematics, University of Georgia, Athens, GA address: pollack@uga.edu