10e2.pdf, a proof of the theorem :
Every nontrivial odd harmonic number must be divisible by a prime greater than 102.
In this article, we prove the theorem without a computer.
A positive integernis said to be harmonic if the harmonic mean of its positive divisors
H(n) =nτ(n) σ(n)
is an integer, where τ(n) and σ(n) denote the number of and the sum of positive divisors of n, respectively. Sinceτ and σare multiplicative, we can express H(n) as
H(n) = Yr
i=1
peiiτ(peii) σ(peii) when the canonical factorization ofnisQr
i=1peii. Using the factsτ(pe) =e+ 1 and σ(pe) = (pe+1−1)/(p−1), we have
(1) H(n)
Yr
i=1
Y
d6= 1, d|(ei+ 1)
Φd(pi) = Yr
i=1
peii(ei+ 1),
where Φd(x) denotes the d-th cyclotomic polynomial. The following facts about harmonic numbers are due to Ore or Garcia. In this article, pe k n means that pe|nandpe+1-n.
Lemma 1 ([4]). Suppose that nis a squarefree harmonic number. Then n= 1 or 6.
Lemma 2 ([2]). Let n be an odd harmonic number and pe k n. Then pe ≡ 1 (mod 4).
In this article, the symbols a, d represent integers greater than 1, andp, pi, q, k primes. Ifk-a, we denote by ordk(a) the order ofa∈(Z/kZ)×.
Lemma 3([3],[5]). A primekdivides Φd(a)if and only ifd=kcordk(a)for some non-negative integerc. Furthermore, ifk|Φd(a) andk|d, thenkkΦd(a).
From Lemma 3, we immediately obtain the fact: if k | Φd(a), then k | d or k≡1 (modd). In the former (resp. latter) case, we say thatk isintrinsic(resp.
primitive).
Lemma 4([1]). A cyclotomic numberΦd(a)has no primitive prime factors if and only ifd= 2, a= 2e−1, ord= 6, a= 2.
Lemma 5. Suppose thatkis a primitive prime factor ofΦq(p). Letgbe a generator of the cyclic group (Z/kmZ)× and putt= (k−1)/q. Thenkm|Φq(p)if and only if pis belong to the subgroup hgtkm−1iof (Z/kmZ)×.
Proof. Note thatkis odd andtis integral sincek≡1 (modq). Ifp≡1 (modk), then Φq(p)≡q (mod k), a contradiction. Hence we easily see that
km|Φq(p) ⇐⇒ km|(p−1)Φq(p) ⇐⇒ pq≡1 (mod km)
⇐⇒ the order ofp∈(Z/kmZ)× is equal toq ⇐⇒ p∈ hgtkm−1i,
1
2
as required. ¤
Proposition 6. Let n (= Qr
i=1peii) be a harmonic number greater than 6, and p the largest prime divisor of n. Then the largest prime divisor of the product Qr
i=1(ei+ 1) is less thanp/2.
Proof. Letqbe a largest prime divisor ofQr
i=1(ei+ 1). Then the left-hand side of (1) is divisible by Φq(pi) for somepi. Note thatq >2 from Lemma 1. By Lemma 4, Φq(pi) has a primitive prime factor k, so that k ≡ 1 (modq). Assume that q≥p/2. Thenk≥2q+ 1≥p+ 1. The left-hand side of (1) is divisible byk, which
is greater than max(p, q), a contradiction. ¤
Lemma 7. Suppose that 3 ≤ p≤97,3 ≤q ≤47, k < 102 and km k Φq(p). All such prime powers km are given by the table below. In particular, if 3 ≤p ≤97 andq= 17,19,31,37,43,47, thenΦq(p)has no prime factors less than 102.
km||Φq(p) Φ3(p) Φ5(p) Φ7(p) Φ11(p) Φ13(p) Φ23(p) Φ29(p) Φ41(p)
p= 3 13 112 23 47 59 83
p= 5 31 11,71 59
p= 7 3,19 29 47 59 83
p= 11 7,19 5 43 83
p= 13 3,61 23 53
p= 17 47 59 83
p= 19 3 59
p= 23 7,79 29 11 83
p= 29 13,67 7 23 59 83
p= 31 3 5,11 23 83
p= 37 3,7,67 11,41 71 47 83
p= 41 5 43 23 59 83
p= 43 3 7
p= 47 37,61 11,31 43 53 23
p= 53 7 11 29 13 472 592
p= 59 11,41 43 23,67 47 29 83
p= 61 3,13,97 5 47 83
p= 67 3,72,31 11,89 79
p= 71 5,11 7 23 472 59
p= 73 3 23
p= 79 3,72,43 13 47 59
p= 83 19 29 47 41
p= 89 11,67 53,79 47
p= 97 3 11,31 43 89 53,79 47
Proof. This fact is easily checked by a computer, however, we give an indirect calculation which can be done without a computer.
Ifk is an intrinsic prime factor of Φq(p), thenq=k andp≡1 (modq). So all cyclotomic numbers Φq(p) with an intrinsic prime factork are given by the table below. In these cases, we havekkΦq(p).
q(=k) p≡1 (modq)
3 7,13,19,31,37,43,61,67,73,79,97 5 11,31,41,61,71
7 29,43,71 11 23,67,89 13 53,79
23 47
29 59
41 83
If k is a primitive prime factor of Φq(p), then k ≡ 1 (modq). For example, suppose thatq= 3 andk= 7. The generators of (Z/7Z)×,(Z/72Z)×and (Z/73Z)× are 3,3 and 3, respectively. Lemma 5 implies that 7m | Φ3(p) if and only if p∈ h32·7m−1i ⊂(Z/7mZ)×. All such primes pare given by the table below.
3
m p
1 11,23,37,53,67,79 2 67,79
3 none
In this way, we have the following table. The mark(m= 2)means thatk2kΦq(p).
In other cases,kkΦq(p).
q k≡1 (modq) p
3 7 11,23,37,53,67(m=2),79(m=2)
3 13 3,29,61
3 19 7,11,83
3 31 5,67
3 37 47
3 43 79
3 61 13,47
3 67 29,37
3 79 23
3 97 61
5 11 3(m=2),5,31,37,47,53,59,71,97
5 31 47,97
5 41 37,59
5 71 5
7 29 7,23,53,83
7 43 11,41,47,59,97
7 71 37
11 23 3,13,29,31,41,59,71,73
11 67 59,89
11 89 67,97
13 53 13,47,89,97
13 79 67,89,97
23 47 3,7,17,37,53(m=2),59,61,71(m=2),79,83,89,97 29 59 3,5,7,17,19,29,41,53(m=2),71,79
41 83 3,7,11,17,23,29,31,37,41,59,61
From the above two cases, we have the lemma. ¤
Lemma 8. Suppose that 3≤p≤97, 3≤q≤47, and Φq(p)has no prime divisors greater than 102. Then (p, q)is one of the following:
(3,3),(3,5),(5,3),(5,5),(7,3),(11,3), (13,3),(23,3),(29,3), (37,3),(47,3),(61,3),(67,3),(79,3).
Proof. We can easily see that Φq(p)>104 follows from one of the following three conditions:
• q= 5 andp≥11,
• q= 7 andp≥7,
• q≥11 andp≥3.
In these cases, Φq(p) has at most two prime factors less than 102 by Lemma 7, hence Φq(p) has a prime factor greater than 102. In the case that (p, q) = (3,7),(5,7) or (7,5), the number Φq(p) has no prime factors less than 102. If q = 3 and p ≥ 11, then it follows that Φ3(p) > 102. From Lemma 7, for any p∈ {17,19,31,41,43,53,59,71,73,83,89,97}, the number Φ3(p) has at most one prime factor less than 102, so it has a prime factor greater than 102. In the other cases, we can see that Φq(p) has no prime divisors greater than 102 by direct cal-
culation. ¤
Proof of the theorem. Suppose thatn(=Qr
i=1peii) is an odd harmonic number and pi<102. From Proposition 6, it follows that the largest prime factor of the product Qr
i=1(ei+ 1) is less than 50, hence the right-hand side of (1) has no prime factors
4
greater than 102, and so does the left-hand side. Therefore, it is necessary that the largest prime factor ofQr
i=1(ei+ 1) is at most 5, from Lemma 8.
If pi = 19, thenei+ 1 is odd by Lemma 2. Since ei+ 1 has no prime factors greater than 50, the left-hand side of (1) has a prime factor greater than 102 by Lemma 8, a contradiction. In this way, we have
pi∈ {/ 19,31,43,59,71,83}.
The right-hand side of (1) is not divided by any of these six primes. Hence we have pi∈ {/ 7,11,67,79}
(see Lemma 8 and the table in the statement of Lemma 7). Continuing these procedure, we have
pi∈ {/ 23,37,41,97}, therefore
pi∈ {/ 47,61,73}, furthermore
pi∈ {/ 13}, and finally,
pi∈ {/ 3}.
So it is necessary that pi ∈ {5,17,29,53,89}. Assume that 5 | (ei+ 1). Then it is necessary that pi = 5 from Lemma 8. The left-hand side of (1) is divided by Φ5(5) = 112, a contradiction. Similarly, we have 3-(ei+1). Assume that 2|(ei+1).
Then the left-hand side of (1) is divided by 3, a contradiction. Therefore, ei+ 1
has no prime factors, and it follows thatn= 1. ¤
References
[1] G. D. Birkhoff and H. S. Vandiver,On the integral divisors ofan−bn, Ann. of Math.
(2)5(1904), 173–180.
[2] M. Garcia,On numbers with integral harmonic mean, Amer. Math. Monthly,61(1954), 89–96.
[3] T. Nagell,Introduction to Number Theory, second ed., Chelsea, New York, 1964.
[4] O. Ore,On the averages of the divisors of a number, Amer. Math. Monthly, 55(1948), 615–619.
[5] H. N. Shapiro,Introduction to the Theory of Numbers, Wiley, New York, 1983.
