On the difference in values of the Euler totient function near prime arguments
aa r X i v : . [ m a t h . N T ] J u l ON THE DIFFERENCE IN VALUES OF THE EULER TOTIENTFUNCTION NEAR PRIME ARGUMENTS
STEPHAN RAMON GARCIA AND FLORIAN LUCA
Abstract.
We prove unconditionally that for each ℓ >
1, the difference ϕ ( p − ℓ ) − ϕ ( p + ℓ ) is positive for 50% of odd primes p and negative for 50%. Introduction
In what follows, p always denotes an odd prime number. The inequality ϕ ( p − > ϕ ( p + 1) appears to hold for an overwhelming majority of twin primes p, p + 2,and to be reversed for small, but positive, proportion of the twin primes [4]. To bemore specific, if the Bateman–Horn conjecture is true, then the inequality aboveholds for at least 65.13% of twin prime pairs and is reversed for at least 0 .
47% ofpairs. Numerical evidence suggests, in fact, that the ratio is something like 98% to2%. In other words, for an overwhelming majority of twin prime pairs p, p + 2, itappears that the first prime has more primitive roots than does the second.Based upon numerical evidence, it was conjectured in [4] that this bias disappearsif only p is assumed to be prime. That is, ϕ ( p − > ϕ ( p + 1) for 50% of primes andthe inequality is reversed for 50% of primes. We prove this unconditionally and,moreover, we are able to handle wider spacings as well. If all primality assumptionsare dropped, then it is known that ϕ ( n − > ϕ ( n + 1) asymptotically 50% of thetime. This follows from work of Shapiro, who considered the distribution functionof ϕ ( n ) /ϕ ( n −
1) [11].Let π ( x ) denote the number of primes at most x and let ∼ denote asymptoticequivalence. The Prime Number Theorem ensures that π ( x ) ∼ x/ log x . Our maintheorem is the following. Theorem 1.1.
Let ℓ be a positive integer. As x → ∞ we have the following:(a) { p x : ϕ ( p − ℓ ) > ϕ ( p + ℓ ) } ∼ π ( x ) .(b) { p x : ϕ ( p − ℓ ) < ϕ ( p + ℓ ) } ∼ π ( x ) .(c) { p x : ϕ ( p − ℓ ) = ϕ ( p + ℓ ) } = o ( π ( x )) . Mathematics Subject Classification.
Key words and phrases.
Euler totient, prime, twin prime, Bateman–Horn conjecture, TwinPrime Conjecture, Brun Sieve.SRG supported by NSF grant DMS-1265973, a David L. Hirsch III and Susan H. HirschResearch Initiation Grant, and the Budapest Semesters in Mathematics (BSM) Director’s Math-ematician in Residence (DMiR) program.F. L. was supported in part by grants CPRR160325161141 and an A-rated researcher awardboth from the NRF of South Africa and by grant no. 17-02804S of the Czech Granting Agency. ℓ ℓ ℓ ℓ ℓ ℓ ℓ ℓ
11 107 19 104 27 357 35 3 43 121 51 371 59 1274 58 12 214 20 3 28 17 36 374 44 39 52 38 60 5385 5 13 98 21 403 29 117 37 97 45 486 53 126 61 1266 231 14 7 22 52 30 507 38 45 46 47 54 303 62 457 43
15 108772
23 136 31 98 39 380 47 124 55 2
63 22654
Table 1.
The number ( p , , ,
743 (the hundred millionthprime) for which ϕ ( p − ℓ ) = ϕ ( p + ℓ ). This number is exceptionally large if ℓ = 4 n −
1; see Theorem 3.1 for an explanation.
A curious phenomenon occurs in (c), in the sense that the decay rate relative to π ( x ) depends upon ℓ in a peculiar manner. Theorem 3.1 shows that (cid:8) p x : ϕ ( p − ℓ ) = ϕ ( p + ℓ ) (cid:9) ≪ x (log x ) if ℓ = 4 n − ,xe (log x ) / otherwise . This does not appear to be an artifact of the proof since it is borne out in numericalcomputations (see Table 1) and is consistent with the Bateman–Horn conjecture.We first prove Theorem 1.1 in the case ℓ = 1. This is undertaken in Section2 and it comprises the bulk of this article. For the sake of readability, we breakthe proof into several steps which we hope are easy to follow. In Section 3, weoutline the modifications necessary to treat the case ℓ >
2. This approach permitsus to focus on the main ingredients that are common to both cases, without gettingsidetracked by all of the adjustments necessary to handle the general case.2.
Proof of Theorem 1.1 for ℓ = 12.1. The case of equality.
Our first job is to show that the set of primes p forwhich ϕ ( p −
1) = ϕ ( p + 1) has a counting function that is o ( π ( x )). We need thefollowing lemma, which generalizes earlier work by Erd˝os, Pomerance, and S´ark¨ozy[3] in the case k = 1. The upper bound (b) in the following was strengthened in apreprint of Yamada [14]. Lemma 2.1 (Graham–Holt–Pomerance [6]) . Suppose that j and j + k have thesame prime factors. Let g = gcd( j, j + k ) and suppose that jtg + 1 and ( j + k ) tg + 1 (2.2) are primes that do not divide j .(a) Then n = j (cid:18) ( j + k ) tg + 1 (cid:19) satisfies ϕ ( n ) = ϕ ( n + k ) .(b) For k fixed and sufficiently large x , the number of solutions n x to ϕ ( n ) = ϕ ( n + k ) that are not of the form above is less than x/ exp((log x ) / ) . HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 3
Consider the case k = 2 and n = p −
1, in which p is prime. Suppose that j and j + 2 have the same prime factors and let g = gcd( j, j + 2). Let us also supposethat t is a positive integer such that jtg + 1 and ( j + 2) tg + 1are primes and p − j (cid:18) ( j + 2) tg + 1 (cid:19) . Since j and j + 2 have the same prime factors, they are both powers of 2. Then j = 2 and j + 2 = 4, so g = 2. Consequently, t + 1 , t + 1 , and 4 t + 3 (2.3)are prime. Reduction modulo 3 reveals that at least one of them is a multiple of 3.The only prime triples produced by (2.3) are (2 , ,
7) and (3 , , r = 1and r = 2, respectively. Consequently, (cid:8) p x : ϕ ( p −
1) = ϕ ( p + 1) (cid:9) < x exp((log x ) / ) + 2 = o ( π ( x )) . (2.4)This is Theorem 1.1.c in the case ℓ = 1.2.2. A comparison lemma.
Instead of comparing ϕ ( p −
1) and ϕ ( p + 1) directly,it is more convenient to compare the related quantities ϕ ( p − p − Y q | ( p − (cid:18) − q (cid:19) and ϕ ( p + 1) p + 1 = Y q | ( p +1) (cid:18) − q (cid:19) , (2.5)in which q is prime. Let S ( p ) := ϕ ( p − p − − ϕ ( p + 1) p + 1 , (2.6)which we claim is nonzero for p >
5. Let P ( n ) denote the largest prime factor of n . Since ϕ ( n ) n = Y q | n (cid:18) q − q (cid:19) , (2.7)it follows that P ( n ) is the largest prime factor of the denominator of ϕ ( n ) /n . Sincegcd( p − , p + 1) = 2, the condition S ( p ) = 0 implies that p − p + 1 are bothpowers of 2. Thus, S ( p ) = 0 holds only for p = 3.Something similar to the following lemma is in [4], although there it was assumedthat p + 2 is also prime. The adjustment for ℓ > Lemma 2.8 (Comparison Lemma) . The set of primes p for which ϕ ( p − − ϕ ( p +1) and S ( p ) have the same sign has counting function asymptotic to π ( x ) as x → ∞ .Proof. In light of (2.4), it suffices to show that ϕ ( p − > ϕ ( p + 1) ⇐⇒ ϕ ( p − p − > ϕ ( p + 1) p + 1 (2.9)on a set of full density in the primes. The forward direction is clear, so we focus onthe reverse. If the inequality on the right-hand side of (2.9) holds, then0 < p (cid:0) ϕ ( p − − ϕ ( p + 1) (cid:1) + ϕ ( p −
1) + ϕ ( p + 1) p (cid:0) ϕ ( p − − ϕ ( p + 1) (cid:1) + ( p −
1) + ( p + 1) S.R. GARCIA AND F. LUCA = p (cid:0) ϕ ( p − − ϕ ( p + 1) + 1 (cid:1) (2.10)because p − p + 1 are even. Since ϕ ( n ) is even for n >
3, it follows that ϕ ( p − − ϕ ( p + 1) >
0. Now appeal to (2.4) to see that strict inequality holds ona set of full density in the primes. (cid:3)
Some preliminaries.
In our later study of the quantity S ( p ), we need toavoid four classes of inconvenient primes. To make the required estimates, we needsome notation. Let x be large, let y := log log x , and define L y := lcm { m : m y } . (2.11)Then L y = e ψ ( y ) , in which ψ ( y ) := X p k y log p is Chebyshev’s function. Since the Prime Number Theorem asserts that ψ ( y ) = y + o ( y ) as y → ∞ , we obtain L y = e y + o ( y ) < e y = (log x ) (2.12)for sufficiently large x . For a positive integer n , let D y ( n ) denote the largest divisorof n that is y -smooth : D y ( n ) := max (cid:8) d : d | n and P ( d ) y (cid:9) . (2.13)On occasion, we will need the Brun sieve. Let f , f , . . . , f m be a collection ofdistinct irreducible polynomials with positive leading coefficients. An integer n is prime generating for this collection if each f ( n ) , f ( n ) , . . . , f m ( n ) is prime. Let G ( x ) denote the number of prime-generating integers at most x and suppose that f = f f · · · f m does not vanish identically modulo any prime. As x → ∞ , G ( x ) ≪ x (log x ) m , in which the implied constant depends only upon m and Q mi =1 deg f i [13, Thm. 3,Sect. I.4.2]. The upper bound obtained in this manner has the same order ofmagnitude as the prediction furnished by the Bateman–Horn conjecture [1].2.4. Inconvenient primes of Type 1.
Let E ( x ) := (cid:8) p x : D y ( p − ∤ L y (cid:9) . We will prove that E ( x ) ≪ xy / log x = o ( π ( x )) . (2.14)Suppose that p ∈ E ( x ). Then (2.13) and the definition of E ( x ) yield a divisor d of p − P ( d ) y and d ∤ L y . These conditions provide a prime power q b with least exponent b such that q b | d, q = P ( q b ) y, and y < q b . (2.15)Indeed, if every y -smooth prime power q b that divides d satisfies q b y , then(2.11) would imply that d | L y , a contradiction. We also observe that the second twoconditions in (2.15) ensure that b > x either p − p + 1 has a prime power divisor q c with c > y/ , y ]. Since p − p − p + 1) and gcd( p − , p + 1) = 2 , HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 5 it follows that d/ (2 , d ) divides one of p + 1, p −
1. There are two cases to consider. • If q = 2, then 2 b − divides one of p − p + 1. Since we aim to prove (2.14) as x → ∞ , we may assume that b > b = 2, the third statement in (2.15)implies log log x <
4. Next observe that (2.15) implies that y/ < b − . Theminimality of b in (2.15) ensures that 2 b − y < y . Thus, 2 b − ∈ [ y , y ] has b − > b > p − p + 1. • If q is odd, then q b divides p − p + 1. The minimality of b ensures that q b − y and the second statement in (2.15) yields y < y < q b = q b − q y , and hence q b ∈ [ y , y ] with b > . For large x , we conclude that one of p − p + 1 has a prime power divisor q c with c > y/ , y ].Let π s ( x ) denote the number of prime powers p a with a > x .Since p a x with a > a = 2 and p x / , or a ∈ [3 , (log x/ log 2)]and p x / , the Prime Number Theorem implies that π s ( x ) = π ( √ x ) + O (cid:0) π ( x / ) log x (cid:1) = (2 + o (1)) x / log x as x → ∞ . Let π ( x ; m, k ) denote the number of primes at most x that are congruentto k modulo m . Then the Brun sieve implies E ( x ) X q b ∈ [ y/ ,y ] b > π ( x ; q b ,
1) + X q b ∈ [ y/ ,y ] b > π ( x ; q b , − ≪ X q b ∈ [ y/ ,y ] b > xϕ ( q b ) log x ≪ X q b ∈ [ y/ ,y ] b > xq b log x ≪ x log x X q b ∈ [ y/ ,y ] b > q b ≪ x log x Z y y/ dπ s ( t ) t x log x Z ∞ y/ dπ s ( t ) t ≪ x log x π s ( t ) t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ y/ + Z ∞ y/ π s ( t ) dtt ≪ x log x ( y/ / ( y/
2) log( y/
2) + Z ∞ y/ t − / (log t ) − / dt ! ≪ xy / (log y )(log x ) = o ( π ( x ))as x → ∞ . This is the desired estimate (2.14).2.5. Inconvenient primes of Type 2.
Fix a large x and define the function h y ( n ) := X r | nr>y r , S.R. GARCIA AND F. LUCA in which r is prime. Let E ( x ) := (cid:26) p x : h y ( p − > y √ log y or h y ( p + 1) > y √ log y (cid:27) . We claim that E ( x ) ≪ π ( x ) √ log y = o ( π ( x )) (2.16)as x → ∞ . This will follow from an averaging argument similar to [7, Lem. 3].The Brun sieve with f ( t ) = rt ± π ( x ; r, ± ≪ π ( x ) r for y r (log x ) uniformly for r in the specified range [13, Thm. 3, Sect. I.4.2]. We use the trivialestimate π ( x ; r, ± xr for (log x ) r x. We also require the upper bound X r > y r = Z ∞ y dπ ( t ) t = π ( t ) t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ y + 2 Z ∞ y π ( t ) dtt ≪ π ( y ) y ≪ y log y , (2.17)which is afforded by the Prime Number Theorem. As x → ∞ , we have X p x h y ( p ±
1) = X p x X r | ( p ± r>y r = X y r x r X p xr | ( p ± X y r x π ( x ; r, ∓ r ≪ X y r (log x ) π ( x ) r + X (log x )
1) + X p xh y ( p +1) > y √ log y h y ( p + 1) X p x h y ( p −
1) + X p x h y ( p + 1) ≪ π ( x ) y log y , which implies (2.16). HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 7
Inconvenient primes of Type 3.
Let ω ( n ) denote the number of distinctprime factors of n and ω y ( n ) the number of distinct prime factors q y of n . Define E ( x ) = (cid:8) p x : ω y ( p − / ∈ [1 . y, . y ] (cid:9) . We claim that E ( x ) ≪ π ( x )log log y = o ( π ( x )) (2.18)as x → ∞ . Although this is essentially a result of Erd˝os [2], we sketch a simplerproof that is easily generalized since we later need to handle p − ℓ instead of p − p ∈ E ( x ), then for large x we have ω y ( p −
1) + 1 = ω y ( p −
1) + ω y ( p + 1)because gcd( p − , p + 1) = 2. Thus, eithermin { ω y ( p − , ω y ( p + 1) } .
75 log log y + 1 . y or max { ω y ( p − , ω y ( p + 1) > .
25 log log y > . y for sufficiently large x . Without loss of generality, we may suppose that ω y ( p − . y or ω y ( p − > . y. (2.19)Then 0 . y ) (cid:0) ω y ( p − − log log y ) (2.20)and similarly if p + 1 occurs in (2.19).We next require the following “Tur´an–Kubilius”-type result; see [8, Lem. 2], [10, § V.5, 1, p. 159]. To study ϕ ( p ± ℓ ) for ℓ = 1 requires a slight generalization; seeLemma 3.2 in Section 3 for a statement and sketch of the proof. Lemma 2.21 (Motohashi) . X p x ( ω y ( p ± − log log y ) = O ( π ( x ) log log y ) . Now return to (2.20), apply Lemma 2.21, and conclude that0 . y ) E ( x ) X p ∈E ( x ) ( ω y ( p − − log log y ) + ( ω y ( p + 1) − log log y ) X p x ( ω y ( p − − log log y ) + ( ω y ( p + 1) − log log y ) = O ( π ( x ) log log y ) . This yields the desired estimate (2.18).2.7.
Inconvenient primes of Type 4.
Let E ( x ) = ( p x : p − ϕ ( p − > (log y ) / ) . (2.22)We claim that E ( x ) ≪ π ( x )(log y ) / = o ( π ( x )) (2.23)as x → ∞ . Since ϕ ( p − ϕ ( p + 1) ϕ ( p − p − ϕ ( p − > (log y ) / implies that p + 1 ϕ ( p + 1) > (log y ) / or p − ϕ ( p − > (log y ) / . A standard application of the Siegel–Walfisz theorem yields X p x p − ϕ ( p − ≪ π ( x ); (2.24)see [10, § I.28, 1b, p 30] or [9]. The same holds with p − p + 1; theadjustments necessary to handle p ± ℓ are discussed in Section 3. Thus, E ( x )(log y ) / X p ∈E ( x ) (cid:18) p − ϕ ( p −
1) + p + 1 ϕ ( p + 1) (cid:19) X p x (cid:18) p − ϕ ( p −
1) + p + 1 ϕ ( p + 1) (cid:19) ≪ π ( x ) , which yields (2.23).2.8. Convenient primes.
Throughout the remainder of the proof, we let 5 p x , in which x is large, and we suppose that p / ∈ E ( x ) ∪ E ( x ) ∪ E ( x ) ∪ E ( x ) . We say that such a prime is convenient . Because gcd( p − , p + 1) = 2, we have D y ( p −
1) = m m , (2.25)in which p − m n , p + 1 = m n , gcd( m , m ) = 2 , (2.26)every prime factor of m m is at most y , and every prime factor of n n is greaterthan y . In particular, gcd( m , n ) = gcd( m , n ) = 1.We claim that ϕ ( m ) m = ϕ ( m ) m (2.27)for sufficiently large x . In light of (2.7), it follows that P ( n ) is the largest primefactor of the denominator of ϕ ( n ) /n . If ϕ ( m ) /m = ϕ ( m ) /m , then P ( m ) = P ( m ) = 2 since gcd( m , m ) = 2. Thus, m and m are powers of 2 and1 = ω ( m ) + ω ( m ) − ω ( m m ) = ω ( D y ( p − ω y ( p − ∈ (cid:2) . y, . y (cid:3) because p / ∈ E ( x ). This is a contradiction for x > .For convenient p x , we have S ( p ) = ϕ ( p − p − − ϕ ( p + 1) p + 1 = ϕ ( m ) ϕ ( n ) m n − ϕ ( m ) ϕ ( n ) m n . We note that S ( p ) = 0 because otherwise P ( p −
1) = P ( p + 1) by (2.7). Sincegcd( p − , p +1) = 2, it would follow that p − p +1 are powers of 2, which occursonly for p = 3. Lemma 2.8 ensures that S ( p ) has the same sign as ϕ ( p − − ϕ ( p +1)on a set of full density in the primes. HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 9
Since p / ∈ E ( x ), for large x we may use the inequality | t + log(1 − t ) | | t | , for | t | , to obtain ϕ ( n ) n = Y r | ( p − r>y (cid:18) − r (cid:19) = exp X r | ( p − r>y log (cid:18) − r (cid:19) ! = exp − X r | ( p − r>y r + O (cid:18)(cid:18) X r | ( p − r>y r (cid:19) (cid:19)! = exp (cid:16) − h y ( p −
1) + O (cid:0) h y ( p − (cid:1)(cid:17) = 1 + O (cid:18) y √ log y (cid:19) , in which the implied constant in the preceding can be taken to be 2. A similarinequality holds if n is replaced by n . Consequently, S ( p ) = ϕ ( m ) m O (cid:18) y √ log y (cid:19)! − ϕ ( m ) m O (cid:18) y √ log y (cid:19)! = ϕ ( m ) m − ϕ ( m ) m + O (cid:18) y √ log y (cid:19) , (2.28)in which the implied constant can be taken to be 4.2.9. Weird primes.
A convenient prime p x is weird if S ( p ) (cid:18) ϕ ( m ) m − ϕ ( m ) m (cid:19) < S ( p ) and ϕ ( m ) /m − ϕ ( m ) /m have opposite signs (the second factoris nonzero if x is large; see (2.27)). If this occurs, then (2.28) tells us that (cid:12)(cid:12)(cid:12)(cid:12) ϕ ( m ) m − ϕ ( m ) m (cid:12)(cid:12)(cid:12)(cid:12) < y √ log y . (2.29)In general, one expects the sign of S ( p ) to be determined by small primes; that is,those primes at most y . If p is weird, then the primes q > y that divide p − m , m ) of positive integers is weird if24 | m m , m m | L y , gcd( m , m ) = 2 , and (2.29) holds . What is the reason for the appearance of the number 24 in the preceding? If p > p − | ( p − x > , then y > P (24) = 3 and 24 | D y ( p − p for which D y ( p −
1) = m m , it makes sense for us to insist that m m is divisible by 24. Lemma 2.30.
Let y > exp(48 ) and D | L y be a multiple of .(a) The number of pairs ( m , m ) with D = m m and gcd( m , m ) = 2 is ω ( D ) . (b) If ω ( D ) ∈ [1 . y, . y ] and D/ϕ ( D ) (log y ) / , then the numberof weird pairs ( m , m ) with D = m m is ≪ ω ( D ) √ log log y . Proof. (a) In what follows, ν p denotes the p -adic valuation function. Write D = Y q ∈ S q ν q ( D ) , in which S is a set of primes that contains 2, S = ω ( D ), and ν (2) >
3. Sincegcd( m , m ) = 2, it follows that ν q ( m ) = ( ν ( D ) − q = 2 , ν q ( D ) if q > . For each of the ω ( D ) primes in S , there are two possible choices for ν q ( m ). Con-sequently, there are 2 ω ( D ) possible pairs ( m , m ).(b) Let ( m , m ) = (2 n , ν ( D ) − n ) or (2 ν ( D ) − n , n )and ( m ′ , m ′ ) = (2 n ′ , ν ( D ) − n ′ ) or (2 ν ( D ) − n ′ , n ′ )be weird, where n , n , n ′ , n ′ are odd, and let D = m m = m ′ m ′ . Supposetoward a contradiction that n | n ′ and n < n ′ . Then (2.29) says that (cid:18) ϕ ( n ) n (cid:19) = 4 ϕ ( m ) m = 4 ϕ ( m ) m (cid:18) ϕ ( m ) m + O (cid:18) y √ log y (cid:19)(cid:19) = 4 ϕ ( m ) ϕ ( m ) m m + O (cid:18) y √ log y (cid:19) (2.31)= 2 ϕ ( m m ) m m + O (cid:18) y √ log y (cid:19) = 2 ϕ ( D ) D + O (cid:18) y √ log y (cid:19) = 2 ϕ ( D ) D O (cid:18) y (log y ) / (cid:19)! (2.32)since D/ϕ ( D ) (log y ) / . The implied constant in (2.32) is 16, in light of (2.29)and the absorption of 4 ϕ ( m ) /m in (2.31). Similar reasoning yields an analogousexpression for ϕ ( n ′ ) /n ′ , with the same implied constant.Let r be the smallest prime divisor of n ′ /n . Use the inequality1 + s t | s − t | (cid:0) | s | + | t | (cid:1) , | s | , | t | , and the fact that (2.32) holds for n and n ′ to deduce that1 + 1 r < r − (cid:18) r − (cid:19) = 1(1 − /r ) = (cid:18) n ′ ϕ ( n ′ ) · ϕ ( n ) n (cid:19) (16 + 16) y (log y ) / = 1 + 48 y (log y ) / . Since r | D and D | L y , we have r y and hence y (log y ) / < r y. This is a contradiction if y > exp(48 ).Hence, the set of odd components n of the parts m as ( m , m ) ranges overweird pairs has the property that no two divide each other. Identifying n withthe set of its odd prime factors, no two n and n ′ , as subsets, are contained one inanother. Sperner’s theorem from combinatorics and Stirling’s formula ensure thatthe number of such n , and hence the number of such pairs ( m , m ), is (cid:18) ω ( D ) ⌊ ω ( D )2 ⌋ (cid:19) ≪ ω ( D ) p ω ( D ) ≪ ω ( D ) √ log log y . (cid:3) (cid:3) Conclusion.
We have shown that the number of inconvenient primes at most x is o ( π ( x )) and hence they can be safely ignored. Each convenient prime p givesrise to a pair ( m , m ) as in (2.25).Suppose that x is large. Let D be a multiple of 24 with D | L y , Dϕ ( D ) (log y ) / , and ω ( D ) ∈ [1 . y, . y ] . (2.33)We wish to count the primes p x for which D y ( p −
1) = D . Denote this numberby π D ( x ). To complete the proof of Theorem 1.1 in the case ℓ = 1, it suffices to showthat (1 / o (1)) π D ( x ) of primes at most x have S ( p ) > / o (1)) π D ( x )have S ( p ) <
0, and that the implied constant is uniform for all D as above.Choose a pair ( m , m ) such that D = m m and gcd( m , m ) = 2. We want tocount the primes p x such that m | ( p −
1) and m | ( p + 1); that is, such that p ≡ m ) and p ≡ − m ) . (2.34)Apply the Chinese Remainder Theorem with moduli m , m or m , m , depend-ing upon which of m , m is exactly divisible by 2, to see that p belongs to anarithmetic progression a m ,m (mod D/ a m ,m , D/
2) = 1. However,we also want gcd (cid:18) p − m m , L y (cid:19) = 1 . For this, we need to work modulo M D := ( D/ Y r y r. To ensure that (2.34) holds, we do the following: • If r | D , then we then want p ≡ ε + r ν r ( D/ λ (mod r ν r ( D/ ) A collection of sets that does not contain X and Y for which X ( Y is a Sperner family . If S isa Sperner family whose union has a total of n elements, then S (cid:0) n ⌊ n ⌋ (cid:1) [12]. for some λ ∈ { , , . . . , r − } . Here, ε = ± r ν r ( D/ divides m or m , respectively. For each r | D , there are r − λ and hencethere are r − p modulo r ν r ( D/ . • If r ∤ D , then we want p ≡ λ (mod r ) for some λ
6∈ { , , r − } . For each r ∤ D ,there are r − p modulo r .Thus, the number of progressions modulo M D that can contain a prime p for which(2.34) occurs is Y r | D ( r − Y r yr ∤ D ( r − . (2.35)By (2.12), the common modulus of all these progressions satisfies M D L y (log x ) for large x . The Siegel–Walfisz theorem says that the number of primes in eachprogression is asymptotically π ( x ) ϕ ( M D ) + O (cid:16) xe − C √ log x (cid:17) (2.36)for some C >
0. Summing up over the number of progressions (or, more precisely,multiplying the (2.36) by the number of acceptable progressions (2.35)), and usingthe fact that ϕ ( M D ) = ( D/ Q r y ( r − π ( x ) D Y r yr ∤ D (cid:18) r − r − (cid:19) + O (cid:16) xL y e − C √ log x (cid:17) . The count depends on D but not on the pair of divisors ( m , m ) of D . We nowapply Lemma 2.30 and obtain π D ( x ) = 2 ω ( D )+1 π ( x ) D Y r yr ∤ D (cid:18) r − r − (cid:19) + O (2 ω ( D ) xL y e − C √ log x ) . (2.37)Although it is not crucial to our proof, we show in Subsection 2.11 that X D π D ( x ) = π ( x ) (cid:18) O (cid:18) √ log log y (cid:19)(cid:19) , (2.38)where the index D runs over all D for which (2.33) holds, because the computationis of independent interest.The product in (2.37) is less than 1 and bounded below by Y r yr ∤ D (cid:18) r − r − (cid:19) ≫ Y r y (cid:18) r − r − (cid:19) = Y r y (cid:18) r ( r − r − (cid:19) (cid:18) − r (cid:19) = Y r y (cid:18) − r − r − (cid:19) Y r y (cid:18) − r (cid:19) ≫ Y r y (cid:18) − r (cid:19) ≫ (log y ) − (2.39) HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 13 by Mertens’ asymptotic formula [10, § VII.29.1b, p. 259]. Since2 ω ( D ) > . y ≫ (log y ) , we examine the main term in (2.37) and conclude that π D ( x ) ≫ π ( x ) D (log y ) . On the other hand, the error term in (2.37) is O (2 ω ( D ) xL y e − C √ log x ) = O (2 . y x (log x ) e − C √ log x )= O (cid:0) (log x ) (log y ) xe − C √ log x (cid:1) = o (cid:18) π ( x ) D (log y ) (cid:19) = o (cid:18) π D ( x )log y (cid:19) . There is a symmetry between non-weird pairs ( m , m ) with ϕ ( m ) m − ϕ ( m ) m > ϕ ( m ) m − ϕ ( m ) m < m , m ) ( m , m ). Indeed, we could return to (2.26)and insist that m | ( p + 1) and m | ( p −
1) instead. The subsequent asymptoticestimates go through in exactly the same manner. Via this transposition, we obtainan asymptotically equal count between the convenient primes p x correspondingto ( m , m ) and the convenient primes p x corresponding to ( m , m ). If onlynon-weird pairs ( m , m ) are taken into account, for fixed D y ( p −
1) = D thissymmetry gives an asymptotically equal count of convenient primes p x with S ( p ) > S ( p ) < m , m ) with D = m m is ≪ ω ( D ) / √ log log y . As x → ∞ , we see from (2.37) that the number of primesat most x that arise from some weird pair is ≪ ω ( D ) π ( x ) D √ log log y Y r yr ∤ D (cid:18) r − r − (cid:19) = O (cid:18) π D ( x ) √ log log y (cid:19) = o ( π D ( x )) . Recall that the non-weird, convenient primes have full density in the set ofprimes. Of such primes p x , the argument above shows that an asymptoticallyequal amount have S ( p ) > S ( p ) < S ( p ) = 0 only for p = 3).This completes the proof of Theorem 1.1 in the case ℓ = 1. (cid:3) Sanity check.
Before extending the preceding proof to the case ℓ >
2, it ishelpful to perform a quick sanity check. Our goal here is to prove (2.38). In lightof (2.37), it suffices to prove that X D ω ( D )+1 D Y r yr ∤ D (cid:18) r − r − (cid:19) = 1 + o (1) , (2.40)in which the index D runs over all D for which (2.33) holds. In particular, (2.38)holds and the preceding product does not run over r = 2 ,
3. These developmentsseems remarkably fortuitous. Let us provide an independent derivation of (2.40),which will help corroborate some of the fine details in the preceding proof.
First write D = 2 k D , in which k >
3, and sum to obtain (cid:18) X k > k (cid:19)(cid:18) X D odd ω ( D ) D (cid:19) = X D odd ω ( D ) D . Now write D = 3 k D and sum over k > (cid:18) X k > k (cid:19)(cid:18) X gcd( D , ω ( D ) D (cid:19) = X gcd( D , ω ( D ) D . For the rest, we use multiplicativity to say that the sum in (2.40) is Y r y (cid:18) r − r − (cid:19) Y r y (cid:18) r − / ( r − X k > r k (cid:19) . However, this is not strictly correct since the sum above stops at the largest power r b y . Moreover, the sum runs over all D without restrictions such as ω ( D ) ∈ [1 . y, . y ] or D/ϕ ( D ) (log y ) / . We deal with these omissionsshortly. For the time being, let us ignore these restrictions. Then the amountinside the Euler factor is1 + 2( r − r − r − r − r − r − , which cancels with the outside ( r − / ( r − r b , in which b is maximal such that r b y . By extending the sum to infinity we incurred an error of2( r − / ( r − X k > b +1 r k = 2 r b ( r −
3) = O (cid:18) y (cid:19) . For 5 r y , the actual Euler factor is r − r − O (cid:18) r − y (cid:19) = r − r − (cid:18) O (cid:18) y (cid:19)(cid:19) Similar considerations apply for r = 2 ,
3. The total multiplicative error is (cid:18) O (cid:18) y (cid:19)(cid:19) π ( y ) = 1 + O (cid:18) π ( y ) y (cid:19) = 1 + O (cid:18) y (cid:19) . (b) We consider only D such that D/ϕ ( D ) (log y ) / . Let the set of remaining D be denoted D . For D ∈ D , we have1 D y ) / · ϕ ( D ) . Applying this inequality and extending then the sum over all possible D , thepiece of the sum over D is at most ≪ π ( x )(log y ) / X D ω ( D )+1 ϕ ( D ) Y r yr ∤ D (cid:18) r − r − (cid:19) . HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 15
We separate out the power of 2 in D as D = 2 k D with k > (cid:18) ϕ (8) + 1 ϕ (16) + · · · + 1 ϕ (2 k ) + · · · (cid:19) = 2 . Then we separate out a factor of 3 from D writing it as D = 3 k D , gettingan Euler factor corresponding to 3 of2 (cid:18) ϕ (3) + 1 ϕ (9) + · · · + 1 ϕ (3 k ) + · · · (cid:19) = 32 . For the remaining primes r >
5, we form the Euler product getting π ( x )(log y ) / Y r y (cid:18) r − r − (cid:19) y Y r =5 r − / ( r − X b > ϕ ( r b ) . The factor inside the parentheses is1 + 2( r − / ( r −
1) 1( r − − /r ) = 1 + 2 r ( r − r − r − r + 3( r − r − r − r − (cid:18) r − (cid:19) . Multiply this by the outside factor ( r − / ( r −
1) and get (cid:18) r − r − (cid:19) r − / ( r − X b > ϕ ( r b ) = 1 + 2( r − . Taking the product of the factors above over r ∈ [5 , y ], we get a convergentproduct. Consequently, π ( x ) X D | L y , | DD/ϕ ( D ) > (log y ) / ω ( D )+1 D Y yr ∤ D (cid:18) r − r − (cid:19) ≪ π ( x )(log y ) / . (c) We need to consider D with ω ( D ) [1 . y, . y ]. From thepreceding material and (2.39), we have Y r y X b > r − / ( r − X b > r b ≍ Y r y (cid:18) r − r − (cid:19) − ≍ (log y ) . We first deal with D with many prime factors. Consider the multiplicativefunction defined for prime powers r b with r > f ( r ) = 2( r − r − r b . If K := ⌊ . y ⌋ , then X r | D = ⇒ r ∈ [5 ,y ] ω ( D ) >K − f ( D ) X k>K − k ! (cid:18) X r ∈ [5 ,y ] b > f ( r b ) (cid:19) k . We have S = X r ∈ [5 ,y ] b > f ( r b ) = X b > r y (cid:18) r b + 2 r b (cid:18) r − r − − (cid:19)(cid:19) = X r ∈ [2 ,y ] b > r b + O (cid:18) X r > b > r b +1 (cid:19) = 2 log log y + O (1) , in which we have used Mertens’ theorem [10, § VII.28.1b]. In the sum P k>K S k /k !,the ratio of two consecutive terms is S k +1 / ( k + 1)! S k /k ! = Sk + 1 = 2 log log y + O (1)2 . y + O (1) < k > K − x , so the first term dominates. With K ! > ( K/e ) K ,the contribution of D with ω ( D ) > . y is at most ≪ S K − ( K − (cid:18) e log log y + O (1)2 . y + O (1) (cid:19) . y + O (1) ≪ (log y ) c , in which c = 2 . e/ . < .
95. Multiplying this by (see (2.39)) Y r y (cid:18) r − r − (cid:19) ≪ (log y ) − , we obtain π ( x ) X D | L y | Dω ( D ) > . y ω ( D )+1 D Y r yr ∤ D (cid:18) r − r − (cid:19) ≪ π ( x )(log y ) . . We use a similar argument for D with ω ( D ) < . y . In this case, let K := ⌊ . y ⌋ . We have to deal with X D : r | D = ⇒ r ∈ [5 ,y ] ω ( D ) < . y − f ( D ) X k K − k ! S k . For k > K − x , the ratio of any two consecutive terms above is S k +1 / ( k + 1)! S k /k ! = Sk + 1 > y + O (1)1 . y + O (1) > , it follows that the last term dominates. Thus, this sum is at most (cid:18) e log log y + O (1) K − (cid:19) K − = (cid:18) e log log y + O (1)1 . y + O (1) (cid:19) . y + O (1) ≪ (log y ) c , in which where c = 1 . e/ . < .
95. Consequently, the contribution of D with ω ( D ) < . y to the sum defining π D ( x ) is ≪ π D ( x )(log y ) . . Putting everything together we obtain (2.40), which is equivalent to (2.38).
HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 17 Proof of Theorem 1.1 for ℓ > ℓ > ℓ = 1, although thereare a number of minor adjustments that must be made. For example, in Lemma2.30 we assumed that D is a multiple of 24. Elementary considerations reveal thatthe following adjustments are necessary for various values of ℓ :(a) D is coprime to all primes that divide ℓ .(b) D is odd if ℓ is even.(c) D is a multiple of 8 if ℓ is odd.(d) D is a multiple of 3 if and only if ℓ is not.More significant modifications are discussed below.3.1. The case of equality.
We need a variant of the inequality (2.4). The estimateprovided by the following theorem involves two special cases. Numerical evidencestrongly suggests that this distinction is not simply a byproduct of our proof; seeTable 1. If we replace the use of the Brun sieve in what follows with an appeal tothe Bateman–Horn conjecture [1], then the larger of the two upper bounds becomesan asymptotic equivalence if the appropriate constant factor is introduced.
Theorem 3.1.
For each ℓ > , (cid:8) p x : ϕ ( p − ℓ ) = ϕ ( p + ℓ ) (cid:9) ≪ x (log x ) if ℓ = 4 n − ,xe (log x ) / otherwise . Proof.
In Lemma 2.1, let k = 2 ℓ and n = p − ℓ , in which p is prime. Suppose that j and j + 2 ℓ have the same prime factors. Since p = j (cid:18) ( j + 2 ℓ ) tg + 1 (cid:19) + ℓ, (3.1)it follows that j is not divisible by any prime factor of ℓ and hence g = gcd( j, ℓ ) = 2.Thus, j = 2 m and j + 2 ℓ = 2 m + n for some m, n >
1. Then 2 m + n = 2 m + 2 ℓ and ℓ = 2 m − (2 n − . If m >
2, then ℓ is even and (3.1) implies that 2 | p , a contradiction. Thus, the upperbound from Lemma 2.1 applies in this case.If m = 1, then j = 2 and ℓ = 2 n −
1. Then p = 2( ℓ + 1) t + ( ℓ + 2) , q = t + 1 and r = ( ℓ + 1) t + 1and we count t x − ( ℓ +2)2( ℓ +1) ∼ x ℓ for which p, q, r are simultaneously prime. Let f ( t ) = 2 n +1 t + (2 n + 1) , f ( t ) = t + 1 , and f ( t ) = 2 n t + 1 . If n is odd, then f (0) ≡ f (2) ≡ f (1) ≡ . There are three possibilities:(a) If f (0) = 3, then f (0) = 1 is not prime and no prime triples are produced.(b) If f (2) = 3, then for each odd n , at most one prime triple is produced. The only odd n <
99 for which a prime triple arises in this manner are n = 1 , , ,
15, from whichwe obtain the triples (11 , , , , , , , , (c) If f (1) = 3, then n = 1 and only the prime triple (7 , ,
3) is produced.In each case, the upper bound from Lemma 2.1 dominates.If n is even, then none of f , f , f vanish identically modulo any prime. TheBrun sieve says that the number of p x for which p, q, r are prime is O ( x/ (log x ) ),which dominates the estimate from Lemma 2.1. (cid:3) A more general comparison lemma.
The next adjustment that is requiredis an analogue of the comparison lemma (Lemma 2.8). This turns out to be moreinvolved than expected. In fact, we first need a generalization of the “Tur´an–Kubilius”-type result from Lemma 2.21. Since this is a minor variant of an existingresult, we only sketch the proof.
Lemma 3.2.
For each ℓ > , X p x ( ω y ( p ± ℓ ) − log log y ) = O ( π ( x ) log log y ) .Proof. Since y (log x ) , apply the Siegel–Walfisz theorem to obtain X p x ω y ( p ± ℓ ) = π ( x ) log log y + O ( π ( x )) , X p x ω y ( p ± ℓ ) = π ( x )(log log y ) + O ( π ( x ) log log y ) , in which the log log y term arises from an application of Mertens’ theorem [10, § VII.28.1b]. Now expand P p x ( ω y ( p ± ℓ ) − log log y ) and apply the preceding. (cid:3) The direct generalization of Lemma 2.8 for ℓ > ℓ issufficiently large, then the +1 in (2.10) becomes too large for the same argumentto work. The evenness of ϕ ( p − ℓ ) − ϕ ( p + ℓ ) is no longer sufficient to push theargument through. Fortunately, we are able to employ the following lemma instead. Lemma 3.3.
For ℓ, m > , (cid:8) p x : ϕ ( p ± ℓ ) ≡ m ) (cid:9) ∼ π ( x ) . Proof.
Fix ℓ >
1. Let ω ( n ) denote the number of distinct prime divisors of n . Then2 ω ( n ) − | ϕ ( n ) since ϕ ( n ) = Q p a k n p a − ( p − ℓ m + 1 ω ( p ± ℓ ) , (3.4)then 2 ℓ m ω ( p ± ℓ ) − and hence ϕ ( p ± ℓ ) ≡ m ) . Thus, it suffices to show that the set of primes p x for which (3.4) fails has acounting function that is o ( π ( x )). Let x be so large that y = log log x satisfies2 + log ℓ log log y and let E ( x ) = (cid:8) p x : ω ( p − ℓ ) < ℓ or ω ( p + ℓ ) < ℓ (cid:9) Let ω y ( n ) denote the number of distinct prime factors q y of n . If p ∈ E ( x ), then ω y ( p − ℓ ) < log log y or ω y ( p + ℓ ) < log log y, and hence ω y ( p − ℓ ) / ∈ [ log log y, log log y ] or ω y ( p + ℓ ) / ∈ [ log log y, log log y ] . HE EULER TOTIENT FUNCTION NEAR PRIME ARGUMENTS 19
Then (log log y ) (cid:0) ω y ( p − ℓ ) − log log y ) or (log log y ) (cid:0) ω y ( p + ℓ ) − log log y ) . Lemma 3.2 ensures that (log log y ) E ( x ) X p ∈E ( x ) ( ω y ( p − ℓ ) − log log y ) + ( ω y ( p + ℓ ) − log log y ) X p x ( ω y ( p − ℓ ) − log log y ) + ( ω y ( p + ℓ ) − log log y ) = O ( π ( x ) log log y ) . Thus, E ( x ) ≪ π ( x )log log y = o ( π ( x )) . (cid:3) (cid:3) Our replacement for the comparison lemma is the following. Since the excep-tional set is o ( π ( x )), it will not affect the proof of Theorem 1.1 in the case ℓ > Lemma 3.5.
For each ℓ > , the set of primes p for which ϕ ( p − ℓ ) − ϕ ( p + ℓ ) and S ( p ) := ϕ ( p − ℓ ) p − ℓ − ϕ ( p + ℓ ) p + ℓ (3.6) have the same sign has counting function ∼ π ( x ) .Proof. By Theorem 3.1, it suffices to show that the set of primes p for which ϕ ( p − ℓ ) > ϕ ( p + ℓ ) ⇐⇒ ϕ ( p − ℓ ) p − ℓ > ϕ ( p + ℓ ) p + ℓ (3.7)has counting function ∼ π ( x ). The forward implication is straightforward, so wefocus on the reverse. If the inequality on the right-hand side of (3.7) holds, then0 < ( p + ℓ ) ϕ ( p − ℓ ) − ( p − ℓ ) ϕ ( p + ℓ )= p [ ϕ ( p − ℓ ) − ϕ ( p + ℓ )] + ℓϕ ( p − ℓ ) + ℓϕ ( p + ℓ ) p [ ϕ ( p − ℓ ) − ϕ ( p + ℓ )] + ℓ ( p − ℓ ) + ℓ ( p + ℓ )= p [ ϕ ( p − ℓ ) − ϕ ( p + ℓ ) + 2 ℓ ] . Let 2 m > ℓ and apply lemma Lemma 3.3 to conclude that ϕ ( p − ℓ ) − ϕ ( p + ℓ ) > p in a set with counting function ∼ π ( x ). (cid:3) Final ingredient.
The only other ingredient necessary to consider ℓ > Lemma 3.8. X p x p ± ℓϕ ( p ± ℓ ) ≪ π ( x ) .Proof. Let σ ( n ) denote the sum of the divisors of n . Then σ ( n ) n = X d | n d and 6 π < σ ( n ) ϕ ( n ) n , for n > see [10, § I.3.5]. The Siegel–Walfisz theorem provides
C > X p x p ± ℓϕ ( p ± ℓ ) ≪ X p x σ ( p ± ℓ )( p ± ℓ ) = X p x X d | ( p ± ℓ ) d = X d x d X p xp ≡∓ d ) X d x π ( x ; ∓ ℓ, d ) d = X d (log x ) π ( x ; ∓ , d ) d + X (log x ) d x π ( x ; ∓ , d ) d X d (log x ) (cid:18) π ( x ) dϕ ( d ) + O (cid:16) xe − C √ log x (cid:17)(cid:19) + X (log x ) d x xd π ( x ) X d< ∞ dϕ ( d ) + O (cid:16) x (log x ) e − C √ log x (cid:17) + x X (log x ) d< ∞ d ≪ π ( x ) + x (log x ) e − C √ log x + x (log x ) ≪ π ( x ) . (cid:3) (cid:3) Acknowledgment.
We thank the referee for suggestions which improved thequality of our paper.
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Department of Mathematics, Pomona College, 610 N. College Ave., Claremont, CA91711
E-mail address : [email protected] URL : http://pages.pomona.edu/~sg064747 School of Mathematics, University of the Witwatersrand, Private Bag 3, Wits 2050,Johannesburg, South Africa, Max Planck Institute for Mathematics, Vivatsgasse 7,53111 Bonn, Germany, Department of Mathematics, Faculty of Sciences, University ofOstrava, 30 dubna 22, 701 03 Ostrava 1, Czech Republic
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