EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 1, No. 4, 2008, (3-21) ISSN 1307-5543 – www.ejpam.com Voronoï type congruences and its applications Takashi Agoh∗,† Department of Mathematics, Tokyo University of Science, Noda, Chiba 278-8510, Japan Abstract. In this paper, we will first deduce Voronoï type congruences for Bernoulli numbers in the even suffix notation. Continuously, we will apply them to extend very important arithmetic properties (such as von Staudt-Clausen’s and Kummer’s congruences) of these numbers to more general situation. AMS subject classifications: 11A07, 11B68, 11Y1 Key words: Bernoulli numbers, Von Staudt-Clausen’s theorem, Voronoï’s congruence, Kummer’s con- gruence, Irregular primes 1. Introduction The Bernoulli numbers Bm (m≥ 0) are defined by the Taylor expansion t et − 1 = ∞ ∑ m=0 Bm m! tm, |t|< 2π. These numbers may be also defined by the recurrence relation Bm =− 1 m+ 1 m−1 ∑ i=0 � m+ 1 i � Bi , B0 = 1. It is easily seen that B2m+1 = 0 and (−1)m−1B2m > 0 for all m≥ 1. Using the Stirling formula n! ∼ (n/e)n p 2πn, we see asymptotically |B2m| ∼ 4 p πm(m/πe)2m. Further, if 3 ≤ m < n, then |B2m/2m|< |B2n/2n|, and also |B2m|/(2m)λ→∞ as m→∞ for every λ≥ 1. Let m ≥ 2 be even and n ≥ 1. If we write Bm = Nm/Dm (Nm, Dm ∈ Z, Dm > 0) in lowest terms, then the Voronoï congruence can be stated as (am− 1)Nm ≡ mDm n−1 ∑ j=1 (a j)m−1 � a j n � (mod n), (1.1) ∗Email address: agoh_takashi@ma.noda.sut.ac.jp (T. Agoh) †The author was supported in part by a grant of the Ministry of Education, Science and Culture of Japan. http://www.ejpam.com 3 c© 2008 EJPAM All rights reserved. T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 4 where a is a positive integer with (a, n) = 1 and [a j/n] is the greatest integer ≤ a j/n. Apply- ing the Voronoï and his type congruences, we are able to deduce various arithmetical prop- erties of Bernoulli numbers. For surrounding landscape on these congruences, see Porubský’s expository article [14]. The prime factorization of Dm can be explicitly stated by the von Staudt-Clausen theorem which asserts that p− 1 | m if and only if p | Dm for a prime p. Further, we know that Dm is square-free. In contrast with the denominator, much less is known about prime divisors of Nm. One of conspicuous facts is von Staudt’s theorem (although this is commonly called Adams’ theorem) which mentions that, for an odd prime p and an even integer m ≥ 2, if p − 1 - m and pe | m (e ≥ 1), then pe | Nm. If we take account of this result, then the question how to find prime divisors of the numerator of Bm/m necessarily arises. The famous congruence Bm/m≡ Bl/l (mod p) (p a prime ≥ 5) for even integers m, l ≥ 2 such that m ≡ l (mod p − 1) and p − 1 - m is due to von Staudt and Kummer, and it shows that Bm/m has period p − 1. We can further state that if p − 1 - m and m ≡ l (mod ϕ(ps)) (s ≥ 1, ϕ the Euler totient function), then (1− pm−1) Bm m ≡ (1− pl−1) Bl l (mod ps). (1.2) This congruence is nothing but a special case of more general formulas given in Theorem 4.1 below, however it is needless to say that this deeply concerns with the construction of p-adic L function. One can consult more details with the beautiful books by Iwasawa [9] and Washington [17]. An odd prime p is said to be irregular if p divides the class number hp of the cyclotomic field Q(ζp) defined by ζp a primitive p th root of unity. Otherwise, it is said to be regular. These notions were first brought by Kummer in his work on Fermat’s Last Theorem. As an irregularity criterion, Kummer showed that p is irregular if and only if p divides at least one of numerators of the (p − 3)/2 Bernoulli numbers B2, B4, ..., Bp−3. It therefore requires to obtain appropriate congruences for getting residues modulo p of B2k (1≤ k ≤ (p− 3)/2). Amazingly, Jensen [10] proved in 1915 that there are infinitely many irregular primes of the form p ≡ 3 (mod 4). The proof was rather easy and, in fact, only some basic arithmetic properties of Bernoulli numbers were used. However, in spite of various efforts by many mathematicians, it has not yet been shown whether there are infinitely many regular primes. Here we should note that Siegel (1964) proved that, under heuristic assumptions, the density of regular primes among the set of all primes is 1/ p e ≈ 0.6065. In this paper, we will first deduce various Voronoï type congruences by generalizing his original (1.1). Applying these, we will extend important arithmetic properties (including von Studt-Clausen’s congruence) on Bernoulli numbers to more general situation. Further we will discuss specific properties of the numerator of Bm/m, whose prime divisors are all irregular. 2. General discussion on Bernoulli numbers In this section, we shall introduce fundamental theorems which are narrating to us very important arithmetical properties of Bernoulli numbers. T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 5 In what follows, we write Bm and βm = Bm/m (m≥ 2 is even) in lowest terms as follows: Bm = Nm Dm and βm = N ′m D′m , where (Nm, Dm) = (N ′m, D′m) = 1 and Dm, D′m > 0. Concerning divisibility properties of the numerator Nm of Bm, we have the following result which is widely known as Adams’ theorem. However, it would be better to call von Staudt’s theorem according to Slavutskii’s suggestion written in his historical observation [15] on von Staudt’s achievements, because this result was first attributed to von Staudt in 1845 before Adams (1878). Theorem 2.1. Let m ≥ 2 be an even integer. If m = pem0 (e ≥ 1, p an odd prime, p - m0) and p - Dm, then pe | Nm. We see from this theorem that p - Dm if and only if p - D′m. The divisor pe of Nm stated above is called an improper divisor of Bm. If d > 0 divides Nm but not m, then it is called a proper divisor of Bm. We can easily show that p is irregular if and only if p is a proper divisor of some Bernoulli number. It is unknown whether there exists an odd prime p such that Bm ≡ 0 (mod p2) for an even m, 2≤ m≤ p− 3. The next theorem is known as the Euler-MacLaurin summation formula. The proof can be easily performed by equating the coefficients of tm+1 in the power series expansions of both sides of the identity t n−1 ∑ i=0 ei t = t et − 1 (ent − 1). Theorem 2.2. Let m, n≥ 1 and Sm(n) = 1m+ 2m+ · · ·+ (n− 1)m. Then Sm(n) = m+1 ∑ j=1 1 j � m j− 1 � Bm+1− jn j . Suppose that pα ‖ 2i + 1 for p ≥ 5 and i ≥ 2. Thus, we can write as 2i + 1 = pαx (α ≥ 1, p - x). Since pαx −α≥ 5α−α≥ 4, 1 2i+ 1 p2i+1 = 1 x ppαx−α ≡ 0 (mod p4). Also, since ord3(Dm) = 1 for all even m ≥ 2, we know � 32Bm−2 � Dm ≡ 0 (mod 32), which implies from Theorem 2.2 the following Theorem 2.3. If m≥ 2 is even and n≥ 1, then DmSm(n)≡ Nmn (mod n2). T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 6 We now introduce the Voronoï congruence which brings various important properties of Bernoulli numbers in relief. Let a and n be positive integers prime to each other. Suppose that q j and r j , j = 1,2, ..., n− 1, are positive integers satisfying a j = q jn+ r j , 0 < r j < n. Here q j = [a j/n]. By direct calculation of (a j− q jn)m = rm j we obtain (a j)m− rm j = m ∑ i=1 (−1)i−1 � m i � (a j)m−ini � a j n �i . Noting that {r j | j = 1, 2, ..., n− 1}= {1,2, ..., n− 1}, if we sum over j = 1,2, ..., n− 1, then (am− 1)Sm(n) = m ∑ i=1 (−1)i−1ni � m i � n ∑ j=1 (a j)m−i � a j n �i , (2.1) so that (am− 1)Sm(n)≡ mn n−1 ∑ j=1 (a j)m−1 � a j n � (mod n2). Ultimately, from Theorem 2.3 we get the following Voronoï congruence: Theorem 2.4. Let m≥ 2 be even and n≥ 1. If a ≥ 1 satisfies (a, n) = 1, then (am− 1)Nm ≡ mDm n−1 ∑ j=1 (a j)m−1 � a j n � (mod n). The next one is known as the famous von Staudt-Clausen theorem. Theorem 2.5. Let p be a prime, m an even integer ≥ 2 and Zp the ring of p-adic integers. If p− 1 - m, then Bm ∈ Zp. If p− 1 | m, then pBm ∈ Zp, precisely, pBm ≡ p− 1 (mod p). (i) More generally, if p is an odd prime and m= kϕ(ps) (k, s ≥ 1), then pBm ≡ p− 1 (mod ps). (ii) Above congruence (i) describes an explicit shape of the denominator of Bm for an even integer m≥ 2. To be exact, Bm is expressed in the form Bm =ω(m)− ∑ p−1|m 1 p , where ω(m) is an integer uniquely determined depending on m and the sum runs over all the primes p such that p − 1 | m. On the one hand, congruence (ii) asserts that if m = kϕ(ps) (k, s ≥ 1) for an odd prime p, then the above ω(m) satisfies ω(m)≡ 1+ ∑ q−1|m q 6=p 1 q (mod ps−1) T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 7 with the sum taken over all the primes q such that q− 1 | m and q 6= p. The congruences mentioned below are widely called Kummer’s congruences for Bernoulli numbers. However, if we comply Slavutskii’s request written in [15], it should be called von Staudt-Kummer’s congruences, because von Staudt was the first contributor who discovered and investigated in details before Kummer. Theorem 2.6. Let p be an odd prime, m an even integer≥ 2 and s an integer with 1≤ s ≤ m−1. If βi(a) = (ai − 1)βi (i ≥ 1) for a positive integer a with p - a, then βm(a) � β p−1(a)− 1 �s ≡ 0 (mod ps). (i) In particular, if p− 1 - m, then βm � β p−1− 1 �s ≡ 0 (mod ps). (ii) The symbolic notation used in above congruences should be understood that we expand in full the left-hand side of (i) (resp. (ii)) using the binomial theorem and β i(a) (resp.β i) is to be replaced by βi(a) (resp.βi) for all i = m+ c(p− 1), c = 0, 1, ..., s. That is, above (i) is exactly the same as the congruence s ∑ c=0 (−1)c � s c � βm+c(p−1)(a)≡ 0 (mod ps), and also (ii) is the congruence exchanged here formally βm+c(p−1)(a) for βm+c(p−1). Note that congruence (ii) is nothing but a special case of (i). In fact, taking a primitive root γ of p and choosing an integer a ≥ 1 such that a ≡ γps−1 (mod ps), we have ap−1 ≡ γϕ(p s) ≡ 1 (mod ps) by Euler’s theorem. Since p - a and p−1 - m, we also have am ≡ am+c(p−1) (mod ps) for each c and am 6≡ 1 (mod p). So, dividing (i) by am− 1, we can immediately deduce (ii). Here, we would like to comment that a similar type congruence to above (ii) without any restriction on m has been obtained by von Staudt. For its explicit formula, see Slavutskii’s article ((4) in [15]). Further, it should be noted that a sequence βi(a) (i ≥ 1) defined above is well-known to be the sequence of moments of a p-adic measure on the ring Zp. The following theorem discovered by E. Lehmer [13] in 1939 was very important for the irregularity testing of primes for many years. Theorem 2.7. Let p be an odd prime and m be an even integer ≥ 2. Also put Q2(m) = 2m − 1,Q3(m) = 1 2 (3m−1),Q4(m) = 1 2 (2m−1)(2m−1−1) and Q6(m) = 1 2 (6m−1+3m−1+2m−1−1). If p− 1 - m− 2, then Qk(m)βm ≡ ∑ 0 0, Hm(n) =εm(n)βm, Km(n; a) =(am− 1)Hm(n) = εm(n)βm(a), H ′m(n) =mHm(n) = εm(n)Bm, K ′m(n; a) =(am− 1)H ′m(n) = εm(n)(a m− 1)Bm. To deduce various important congruences for composite moduli, we introduce generalized Voronoï type congruences as follows: T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 9 Theorem 3.1. Let m ≥ 2 be even and n ≥ 1. Also, let w and w′ be arbitrary positive multiples of nδν and nν , respectively. For a positive integer a with (a, w) = (a, w′) = 1, we have Km(n; a)≡ w−1 ∑ j=1 ( j,n)=1 (a j)m−1 � a j w � (mod n), (i) K ′m(n; a)≡m w′−1 ∑ j=1 ( j,n)=1 (a j)m−1 � a j w′ � (mod n). (ii) Proof. Since the proofs of (i) and (ii) are almost the same, we shall give below only the proof of (i). First, consider the Voronoï congruence in Theorem 2.4 replaced n by w: (am− 1)Nm ≡ mDm w−1 ∑ j=1 (a j)m−1 � a j w � (mod w). Since (mDm/δν , n) = 1 and w/δν ≡ 0 (mod n), dividing this by mDm we get (am− 1)βm ≡ w−1 ∑ j=1 (a j)m−1 � a j w � (mod n). (3.1) Here the sum on the right-hand side can be expressed as w−1 ∑ j=1 (a j)m−1 � a j w � = w−1 ∑ j=1 ( j,n)=1 (a j)m−1 � a j w � + w−1 ∑ j=1 ( j,n)6=1 (a j)m−1 � a j w � = w−1 ∑ j=1 ( j,n)=1 (a j)m−1 � a j w � − ∑ d|n d>1 µ(d)dm−1    w/d−1 ∑ j=1 (a j)m−1 � a j w/d �    , (3.2) where µ is the Möbius function. Now consider congruence (3.1) replaced w by w/d, and multiply it by µ(d)dm−1. Then we have µ(d)dm−1βm(a)≡ µ(d)dm−1 w/d−1 ∑ j=1 (a j)m−1 � a j w/d � (mod n). Noting that εm(n) = ∑ d|nµ(d)d m−1 (where µ(1) = 1), from (3.1) and (3.2) Km(n; a) = � 1+ ∑ d|n d>1 µ(d)dm−1 � βm(a)≡ w−1 ∑ j=1 ( j,n)=1 (a j)m−1 � a j w � (mod n), which is exactly congruence (i), as desired. T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 10 Theorem 3.2. Let m≥ 2 be even and n≥ 1. Then nH ′m(n)≡ n−1 ∑ j=1 ( j,n)=1 jm (mod n). Proof. Using the similar method to that stated in the proof of Theorem 3.1, we express Sm(n) as Sm(n) = n−1 ∑ j=1 ( j,n)=1 jm+ n−1 ∑ j=1 ( j,n)6=1 jm = n−1 ∑ j=1 ( j,n)=1 jm− ∑ d|n d>1 µ(d)dm �n/d−1 ∑ j=1 jm � . By Theorem 2.5 we see ordp(Dm) ∈ {0,1} for all prime divisors p of n, hence from Theorem 2.2 Sm(n) ≡ nBm (mod n) and Sm(n/d) ≡ (n/d)Bm (mod n/d) for any positive divisor d of n. Multiplying the latter one by µ(d)dm, we have µ(d)dmSm (n/d)≡ µ(d)dm−1nBm (mod n). Substituting these congruences for every d into the above, it follows that nH ′m(n) = � 1+ ∑ d|n d>1 µ(d)dm−1 � nBm ≡ n−1 ∑ j=1 ( j,n)=1 jm (mod n), which is precisely the congruence indicated. With above notations, one can state Corollary 3.3. If ϕ(n) | m, then nH ′m(n)≡ ϕ(n) (mod n). Proof. If ( j, n) = 1 and ϕ(n) | m, then jm ≡ 1 (mod n). So the result follows immediately from Theorem 3.2. Note that congruence (i) in Theorem 2.5 is given as a special case of Corollary 3.3 for the case n= p. Let Zn = ∩p|nZp (n ≥ 2) be the ring of rational numbers which are n-integral. Suppose that m≥ 2 is even and ϕ(n) | m. Writing as Bm =ω(m)+ xm+ ym, where ω(m) is the integer mentioned in Section 2, xm ∈ Zn and ym /∈ Zn, we obviously see ym = − ∑ p|n 1/p. On the one hand, the right-hand side of the congruence in Theorem 3.2 is unchanged modulo n even if we replace m by l ≥ 2 satisfying m ≡ l (mod ϕ(n)). Therefore, if m ≡ l (mod ϕ(n)) for n≥ 3, then we have nH ′m(n)≡ nH ′l (n) (mod n). However, since Bm ∈ Zn if and only if Bl ∈ Zn, this congruence is interesting only in the case Bm /∈ Zn. As recurrence relations for H ′m(n), we can give the following T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 11 Theorem 3.4. Let m≥ 1, n≥ 2 and a ≥ 1 be integers with (a, n) = 1. Then � H ′(n) + n �m+1−H ′m+1(n) = (m+ 1) n−1 ∑ j=1 ( j,n)=1 jm, (i) (am− 1) n � H ′(n) + n �m+1−H ′m+1(n) o (ii) =(m+ 1) m ∑ i=1 (−1)i+1 � m i � ni n−1 ∑ j=1 ( j,n)=1 (a j)m−i � a j n �i . We do not give the proof of this theorem, because both congruences (i) and (ii) are nothing but special cases of known formulas for generalized Bernoulli numbers (see Theorems 1 and 3 in Agoh [1]). In addition, we note that Theorem 3.2 can be also derived from above (i). Now, we would like to introduce an interesting problem related to a characterization of primes, which is deeply related to the von Staudt-Clausen congruence. Put G = {n≥ 2 | Sn−1(n)≡ n− 1 (mod n)}. By Fermat’s little theorem, it is clear that the set G contains any primes. In 1950, Giuga [8] conjectured that G is precisely the same as the set of all the primes. By elementary observation, we see that the following statements are all equivalent (for details, see Agoh [2], Borwein et al. [4] and Kellner [12]): (a) n ∈ G, (b) p2(p− 1) | n− p for any prime divisor p of n, (c) nBn−1 ≡ n− 1 (mod n). Therefore, we can restate Giuga’s conjecture as follows: Conjecture 3.5. Let n≥ 2. Then nBn−1 ≡ n− 1 (mod n) if and only if n is a prime. A composite n is called a Carmichael number if an−1 ≡ 1 (mod n) for every positive integer a with (a, n) = 1. Such the smallest number is 561 = 3 · 11 · 17. It is easily shown that n is a Carmichael number if and only if n = p1p2 · · · pg (the product of some distinct odd primes) and p(p − 1) | n− p for each prime divisor p of n. Further, this number may be character- ized by nBn−1 ≡ −n ∑ p|n 1/p (mod n). In their outstanding paper [3], Alford, Granville and Pomerance proved that there exist infinitely many Carmichael numbers. We can show that if there is a composite number n ∈ G, then n is the Carmichael number and the number of its prime divisors must be greater than the smallest prime divisor of n. Also we can say that if n ∈ G is composite, then (cf. Agoh [2])    pBn−p ≡ p− 1 (mod p3), pB(n/p)−1 ≡ p− 1 (mod p2), pB((n/p)−1)/p ≡ p− 1 (mod p) T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 12 for all prime divisors p of n. It is unknown whether there is a composite number n belonging to G. Recently, it was shown in Borwein et al. [4] that any counter example has at least 12055 digits. For more extensive computations, see Fee and Plouffe [7]. If it is possible to appreciate the justification of the above conjecture, then an interesting characterization of primes can be given by means of the Bernoulli number. 4. Generalized von Staudt-Kummer congruences In this section, we will generalize von Staudt-Kummer congruences stated in Theorem 2.6. In what follows, we assume that n, s, mr , ar , br (r = 1, 2, ..., k) are positive integers with n≥ 3 and (ar , n) = 1. Theorem 4.1. For each r = 1, 2, ..., k + 1, we assume that mr is even and λr ∈ Zn satisfies ∑k+1 i=1 λi ≡ 0(mod n). Then we have k ∏ r=1 Kmr (n; ar) k ∑ r=1 λr K brϕ(n)(n; ar) +λk+1 !s ≡ 0 (mod ns). (i) In particular, if n= pα (α≥ 1, p an odd prime) and p− 1 - mr , r = 1,2, ..., k, then k ∏ r=1 Hmr r (n) k ∑ r=1 λr H brϕ(n) r (n) +λk+1 !s ≡ 0 (mod ns). (ii) The symbolic expression used in the above statement means that we expand the left-hand side of (i) (resp. (ii)) in full by making use of the multinomial theorem regarding the K ’s (resp. H ’s) as ordinary real numbers, and afterwards, K l(n; ar) (resp. H l r(n)) is to be replaced by Kl(n; ar) (resp. Hl(n)) for each r. Proof. Since n ≥ 3, we know that all of mr + cbrϕ(n) for r = 1, 2, ..., k and c = 0, 1, ..., s are even. Also, it follows that p − 1 | mr if and only if p − 1 | mr + cbrϕ(n) for each prime divisor p of n, hence ordp(Dmr ) = ordp(Dmr+cbrϕ(n)). We now put ξ= ns ∏ p|n p fp+gp , where fp = max 1≤r≤k 0≤c≤s ¦ ordp(mr + cbrϕ(n)) © and gp = max 1≤r≤k ¦ ordp(Dmr ) © . Noticing that m and n can be chosen independently, we may consider congruence (i) in The- orem 3.1 which is replaced a, m, n and w by ar , mr + cbrϕ(n), ns and ξ, respectively: Kmr+cbrϕ(n)(n; ar)≡ ξ−1 ∑ jr=1 ( jr ,n)=1 (ar jr) mr+cbrϕ(n)−1 � ar jr ξ � (mod ns), T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 13 which is valid for all r = 1,2, ..., k and c = 0,1, ..., s. Making use of this congruence, we can deduce the following k ∏ r=1 Kmr (n; ar) k ∑ r=1 λr K brϕ(n)(n; ar) +λk+1 !s = ∑ 0≤c1,...,ck+1≤s c1+···+ck+1=s � s c1, ..., ck+1 � k ∏ r=1 λcr r Kmr+cr brϕ(n)(n; ar) ! λ ck+1 k+1 ≡ ∑ 0≤c1,...,ck+1≤s c1+···+ck+1=s � s c1, ..., ck+1 � � k ∏ r=1 � λcr r ξ−1 ∑ jr=1 ( jr ,n)=1 (ar jr) mr+cr brϕ(n)−1 � ar jr ξ � �� λ ck+1 k+1 ≡ ξ−1 ∑ j1=1 ( j1,n)=1 · · · ξ−1 ∑ jk=1 ( jk ,n)=1 ¨ ∑ 0≤c1,...,ck+1≤s c1+···+ck+1=s � s c1, ..., ck+1 � × k ∏ r=1 λcr r (ar jr) mr+cr brϕ(n)−1 � ar jr ξ � ! λ ck+1 k+1 « ≡ ξ−1 ∑ j1=1 ( j1,n)=1 · · · ξ−1 ∑ jk=1 ( jk ,n)=1    k ∑ r=1 λr(ar jr) brϕ(n)+λk+1 !s k ∏ r=1 (ar jr) mr−1 � ar jr ξ �    (mod ns). Since (ar jr , n) = 1 for each r, under the assumption for the λ’s we have k ∑ r=1 λr(ar jr) brϕ(n)+λk+1 ≡ k+1 ∑ i=1 λi ≡ 0 (mod n), which offers congruence (i) as desired. Next, assume that n = pα (α ≥ 1, p an odd prime) and p− 1 - mr (r = 1, 2, ..., k). In this case, n has a primitive root γ, so take a positive integer a with a ≡ γns−1 (mod ns). Then, by Euler’s theorem we see aϕ(n) ≡ γns−1ϕ(n) ≡ γϕ(n s) ≡ 1 (mod ns). Now choose especially a1 = a2 = · · ·= ak = a in (i). Then acϕ(n) ≡ 1 (mod ns) for any c ≥ 0 and (amr − 1, n) = 1 for each r. Dividing (i) by ∏k r=1 (a mr − 1), we can deduce (ii) immediately. With above notations, we have Corollary 4.2. Let p be an odd prime, mr ≥ s+1 for r = 1, 2, ..., k and assume that λi ∈ Zp (i = 1,2, ..., k+ 1) satisfy ∑k+1 i=1 λi ≡ 0(mod p). Then k ∏ r=1 βmr r (ar) k ∑ r=1 λrβ br (p−1) r (ar) +λk+1 !s ≡ 0 (mod ps). (i) T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 14 In particular, if p− 1 - mr (r = 1, 2, ..., k), then k ∏ r=1 βmr r k ∑ r=1 λrβ br (p−1) r +λk+1 !s ≡ 0 (mod ps). (ii) It may be unnecessary to comment on the symbolic notation used in the above corollary, but we want to explain it once more to make doubly sure. Above congruence (i) means that we calculate the left-hand side of (i) in full regarding βr(ar) (r = 1,2, ..., k) as ordinary real numbers, and replace β g(r) r (ar) by βg(r)(ar) for each r and various values of g(r). Hence, (i) is precisely the same as ∑ 0≤c1,...,ck≤s c1+···+ck+1=s � s c1, ..., ck+1 � k ∏ r=1 λcr r βmr+cr br (p−1)(ar) ! λ ck+1 k+1 ≡ 0 (mod ps). We do not say fully, but the symbolic notation used in (ii) should be also understood similarly to that used in (i). Proof. Take n= p in Theorem 4.1. Since mr−1≥ s for each r, we get εmr (ps) = 1−pmr−1 ≡ 1 (mod ps), so that Hmr (ps)(ar)≡ βmr (ar) (mod ps). Additionally, if p−1 - mr (r = 1,2, ..., k), then we can choose a positive integer a with (amr − 1, p) = 1 for each r. Therefore, using the same methods as in the proof of Theorem 4.1, we get the indicated congruences. As a special case of Corollary 4.2, we may state the following Corollary 4.3. Let p be an odd prime, s, b ≥ 1, m be an even integer ≥ 2, m≥ s+1 and assume that λ1,λ2 ∈ Zp satisfy λ1+λ2 ≡ 0 (mod p). If a is a positive integer with (a, p) = 1, then βm(a) � λ1β b(p−1)(a) +λ2 �s ≡ 0 (mod ps). (i) In particular, if p− 1 - m, then βm � λ1β b(p−1)+λ2 �s ≡ 0 (mod ps). (ii) Proof. Take k = 1 in Corollary 4.2. Choosing especially b = λ1 = 1 and λ2 =−1 in the above corollary, we can derive readily Theorem 2.6. Corollary 4.4. Let m be an even integer ≥ 2, n, a be positive integers with (a, n) = 1 and n≥ 3. If m≡ l (mod ϕ(n)) for l ≥ 2, then Km(n; a)≡ Kl(n; a) (mod n). (i) In particular, if p is an odd prime with p−1 - m and m≡ l (mod ϕ(pα)) (α≥ 1) for l ≥ 2, then Hm(p)≡ Hl(p) (mod pα). (ii) T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 15 Proof. Take s = k = λ1 = 1 and λ2 =−1 in Theorem 4.1 and note that Hi(pα) = Hi(p) for any i ≥ 1. Then, congruence (i) and (ii) are immediate. Let n, a be positive integers with n ≥ 3 and (a, n) = 1, and let s, b be positive integers. Then we obtain Theorem 4.5. Let m ≥ 2 be an even integer and assume that λ1,λ2 ∈ Zn satisfy λ1 + λ2 ≡ 0 (mod n). Then K ′m(n; a) � λ1K ′bϕ(n)(n; a) +λ2 �s ≡ 0 (mod ns−1). (i) In particular, if n = pα (α ≥ 1, p an odd prime), p− 1 - m and λ1,λ2 ∈ Zp satisfy λ1 + λ2 ≡ 0 (mod p), then H ′m(n) � λ1H ′bϕ(n)(n) +λ2 �s ≡ 0 (mod ns−1). (ii) Proof. As defined in Section 3, let ν = ∏ p|n pvp (where vp = ordp(Dm)) and put η = nsν . Since ordp(Dm) = ordp(Dm+cbϕ(n)) for each prime divisor p of n and c = 0, 1, ..., s, we may consider the Voronoï type congruence (ii) in Theorem 3.1 replaced m, n and w′ by m+ cbϕ(n), ns and η, respectively. Noting that εm(ns) = εm(n) and hence K ′m(n s; a) = K ′m(n; a) for any m, we have K ′m+cbϕ(n)(n; a)≡ (m+ cbϕ(n)) η−1 ∑ j=1 ( j,n)=1 (a j)m+cbϕ(n)−1 � a j η � (mod ns). Using this congruence, it follows that K ′m(n; a) � λ1K ′bϕ(n)(n; a) +λ2 �s = s ∑ c=0 � s c � λc 1K ′m+cbϕ(n)(n; a)λs−c 2 ≡ s ∑ c=0 � s c � λc 1λ s−c 2 � (m+ cbϕ(n)) η−1 ∑ j=1 ( j,n)=1 (a j)m+cbϕ(n)−1 � a j η � � ≡ s ∑ c=0 � s c � λc 1λ s−c 2 � m η−1 ∑ j=1 ( j,n)=1 (a j)m+cbϕ(n)−1 � a j η � � + s ∑ c=0 � s c � λc 1λ s−c 2 � cbϕ(n) η−1 ∑ j=1 ( j,n)=1 (a j)m+cbϕ(n)−1 � a j η � � ≡m η−1 ∑ j=1 ( j,n)=1 (a j)m−1 s ∑ c=0 � s c � λc 1λ s−c 2 (a j)cbϕ(n) ! � a j η � + sλ1 bϕ(n) η−1 ∑ j=1 ( j,n)=1 (a j)m+bϕ(n)−1 s ∑ c′=0 � s− 1 c′ � λc′ 1 λ s−1−c′ 2 (a j)c ′bϕ(n) ! � a j η � T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 16 ≡m η−1 ∑ j=1 ( j,n)=1 (a j)m−1 � λ1(a j)bϕ(n)+λ2 �s � a j η � + sλ1 bϕ(n) η−1 ∑ j=1 ( j,n)=1 (a j)m+bϕ(n)−1 � λ1(a j)bϕ(n)+λ2 �s−1 � a j η � (mod ns). Since (a j)bϕ(n) ≡ 1 (mod n) if (a j, n) = 1, we have, for κ= s, s− 1, � λ1(a j)bϕ(n)+λ2 �κ ≡ � λ1+λ2 �κ ≡ 0 (mod ns−1). From this congruence we can readily derive (i). When p− 1 - m, (ii) can be easily given from (i) by the same arguments as done for the proof of (ii) in Theorem 4.1. Using the same notations as in Theorem 4.5, we can state Corollary 4.6. Let p be an odd prime and assume that λ1,λ2 ∈ Zp satisfy λ1+λ2 ≡ 0 (mod p). If m≥ s and β ′i (a) = iβi(a) = (ai − 1)Bi (i ≥ 1), then β ′m(a) � λ1β ′b(p−1)(a) +λ2 �s ≡ 0 (mod ps−1). (i) In particular, if p− 1 - m, then Bm � λ1Bb(p−1)+λ2 �s ≡ 0 (mod ps−1). (ii) Proof. Since m≥ s, we see εm(p) = 1− pm−1 ≡ 1 (mod ps−1), and so the corollary follows from Theorem 4.5. To obtain some generalized von Staudt-Kummer congruences in this section, we made use of Voronoï type congruences stated in Section 3. However, besides our method, there are other sagacious ones to accomplish the purpose. Indeed, two sequences Hm(n) and Km(n; a) (m≥ 2, even) defined in Section 3 are the moments of p-adic measures on the unit group Z×p of Zp and this realization brings us the congruences such as Theorem 4.1 (cf., e.g., Young [18, 19]). For the other method obtaining extended Voronoï and von Staudt-Kummer congruences, see Sun’s interesting approach in [16]. 5. Generalization of Lehmer’s congruences In this section, we will extend Lehmer’s congruences in Theorem 2.7 to more general moduli. Assume that m ≥ 2 is even and p− 1 - m− 2 for all prime divisors p of n. Then we have m 6= 2 (hence m ≥ 4) and Bm−2 ∈ Zp. Also it is clear that 2 - n and 3 - n, hence p ≥ 5. Recall the Euler-MacLaurin summation formula (Theorem 2.2) and observe each term on the right-hand side of this formula. For simplicity, we put X j = 1 j � m j− 1 � Bm+1− jn j for j = 1, 2, ..., m+ 1. T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 17 If j is an even integer with m− 2 ≥ j ≥ 2, then X j = 0. So we observe here only the terms X j with j = 2i+ 1 (i = 0,1, ..., m/2) and j = m. Let p be any prime divisor of n. If m > 2i ≥ 4, then ordp(Dm−2i) ∈ {0,1} and p2i−3 ≥ 52i−3 ≥ 2i + 1, hence ordp � X2i+1 � ≥ −(2i − 3) − 1 + (2i + 1) = 3. This gives X2i+1 ≡ 0 (mod n3) for all i ≥ 2. Also, since ordp(Dm−2) = 0 by the assumption, m≥ 4 and (n, 6) = 1, it follows that X3 = 1 3 �m 2 � Bm−2n3 ≡ 0 (mod n3) and Xm = 1 m � m m−1 � B1nm =−1 2 nm ≡ 0 (mod n3). Consequently, we get from Theorem 2.2 Sm(n)≡ Bmn (mod n3). On the other hand, we obtain from (2.1) that if (a, n) = 1 , a ≥ 1, then (am− 1)Sm(n)≡ mn n−1 ∑ j=1 (a j)m−1 � a j n � − m(m− 1) 2 n2 n−1 ∑ j=1 (a j)m−2 � a j n �2 (mod n3). Consequently, we have (am− 1)Bm ≡ m n−1 ∑ j=1 (a j)m−1 � a j n � − m(m− 1) 2 n n−1 ∑ j=1 (a j)m−2 � a j n �2 (mod n2). Let δ be as in Section 2 and set η′ = nδ. Since (a,η′) = 1, we may consider the above congruence replaced n by η′. That is, (am− 1)Bm ≡ m η′−1 ∑ j=1 (a j)m−1 � a j η′ � − m(m− 1) 2 η′ η′−1 ∑ j=1 (a j)m−2 � a j η′ �2 (mod η ′2), which implies (am− 1)βm ≡ η′−1 ∑ j=1 (a j)m−1 � a j η′ � − m− 1 2 η′ η′−1 ∑ j=1 (a j)m−2 � a j η′ �2 (mod n2). (5.1) Putting, for i = 0, 1, ..., a− 1, G(a)i = ¨ j ∈ Z | iη′ a < j < (i+ 1)η′ a « , and U (a)i = ∑ j∈G(a)i jm−1, V (a)i = ∑ j∈G(a)i jm−2, above (5.1) can be expressed as (am− 1)βm ≡ am−1 a−1 ∑ i=1 iU (a)i − m− 1 2 am−2η′ a−1 ∑ i=1 i2V (a)i (mod n2). (5.2) T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 18 On the other hand, since m is even, it follows that U (a)i = ∑ j∈G(a)a−1−i (η′− j)m−1 ≡− ∑ j∈G(a)a−1−i jm−1+ (m− 1)η′ ∑ j∈G(a)a−1−i jm−2 ≡− U (a)a−1−i + (m− 1)η′V (a)a−1−i (mod n2). We also have V (a)i = ∑ j∈G(a)a−1−i (η′− j)m−2 ≡ ∑ j∈G(a)a−1−i jm−2 ≡ V (a)a−1−i (mod n). By Theorem 2.2 we see, since Bm−2 ∈ Zp for any prime divisor p of n such that p− 1 - m− 2, Sm−2(η ′) = a−1 ∑ i=0 V (a)i ≡ 0 (mod n). We are now able to prove the following generalizations of Lehmer’s congruences stated in Theorem 2.7. Theorem 5.1. Let n be an odd integer, m an even integer ≥ 2, η′ = nδ, and let Qk(m) (k = 2,3, 4,6) be as mentioned in Theorem 2.7. If p− 1 - m− 2 for every prime divisor p of n, then Qk(m)βm ≡ ∑ 0< j<η′/k (η′− k j)m−1 (mod n2), k = 2, 3,4, 6, (5.3) provided that every prime divisor p of n is ≥ 7 when k = 6. Proof. Since each proof of (5.3) for k = 2,3, 4,6 is similar, we give only the proof for k = 6, under the assumption that (5.3) holds for k = 2,3. In here and what follows, we write Ui = U (6)i and Vi = V (6)i (i = 0, 1, ..., 5) for simplification. From the congruences mentioned above we get Ui ≡−U5−i+(m−1)η′V5−i (mod n2) and Vi ≡ V5−i (mod n) for i = 3,4, 5. Since (n, 6) = 1, by taking a = 6 in (5.2), we obtain (6m− 1)βm ≡6m−1 5 ∑ i=1 iUi − m− 1 2 6m−2η′ 5 ∑ i=1 i2Vi ≡− 6m−1(5U0+ 3U1+ U2) + m− 1 2 6m−2η′(35V0+ 31V1+ 23V2) (mod n2). Under the given condition, Bm−2 ∈ Zp for every prime divisor p of n, so we get ∑5 i=0 Vi ≡ 2 � V0+ V1+ V2 � ≡ 0 (mod n), which implies V2 ≡ −V0 − V1 (mod n) since n is odd. Conse- quently, dividing the above congruence by 6m−1 we have (6− 61−m)βm ≡−(5U0+ 3U1+ U2) + m− 1 3 η′(3V0+ 2V1) (mod n2). T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 19 Assume that (5.3) holds for k = 2 and 3, i.e., in equivalent forms (2− 21−m)βm ≡− (U0+ U1+ U2) + m− 1 2 η′(V0+ V1+ V2) ≡− (U0+ U1+ U2) (mod n2), (3− 31−m)βm ≡− 2(U0+ U1) + 2(m− 1) 3 η′(V0+ V1) (mod n2). Combining these congruences, we deduce (1+ 21−m+ 31−m− 61−m)βm = ¦ −(2− 21−m)− (3− 31−m) + (6− 61−m) © βm ≡− 2U0+ m− 1 3 η′V0 (mod n2), that is, (6m−1+ 3m−1+ 2m−1− 1)βm ≡2 � −6m−1U0+ (m− 1)6m−2η′V0 � ≡2 ∑ j∈G(6)0 (η′− 6 j)m−1 (mod n2), which completes the proof of (5.3) for k = 6. 6. Properties of the numerator of βm In this final section, we would like to discuss some basic properties of the numerator of βm for an even integer m≥ 2. As defined in Section 2, let N ′m be the numerator of βm. Theorem 6.1. Let n = p1p2 · · · ps be the product of some distinct irregular primes such that (1 2 (pi − 1), 1 2 (p j − 1)) = 1 for every i, j = 1,2, ..., s with i 6= j. Then there exists an even integer m≥ 2 such that N ′m ≡ 0 (mod n). Proof. Since all the primes pi , i = 1,2, ..., s, are irregular, there exist even integers 2mi , 1 ≤ mi ≤ 1 2 (pi − 3), such that B2mi ≡ 0 (mod pi). Since (1 2 (pi − 1), 1 2 (p j − 1)) = 1 if i 6= j, by the Chinese remainder theorem the system of congruences X ≡ mi (mod 1 2 (pi − 1)), i = 1, 2, ..., s, has uniquely the solution X = α > 0 modulo 1 2s ∏s i=1(pi − 1) in common. If we choose m = 2α, then pi − 1 - m and βm ≡ β2mi ≡ 0 (mod pi) by Theorem 2.6, which are valid for all i = 1,2, ..., s. This completes the proof. It is evident that if n is divisible by at least two distinct irregular primes p and q with (1 2 (p − 1), 1 2 (q − 1)) 6= 1, then there does not exist an even integer m ≥ 2 such that N ′m ≡ 0 (mod n). We shall give here an example of Theorem 6.1 for n = 37 · 59 · 131, the product of three irregular primes satisfying the indicated condition. All the irregularity indices of 37,59 and T. Agoh / Eur. J. Pure Appl. Math, 1 (2008), (3-21) 20 131 are equal to 1 and the corresponding irregular pairs (p, 2m) satisfying B2m ≡ 0 (mod p) (1 ≤ m ≤ (p− 3)/2) are (37, 32), (59,44) and (131,22), respectively. So consider the system of congruences X ≡ 16 (mod 18), X ≡ 22 (mod 29), X ≡ 11 (mod 65). Then we find the common solution X ≡ 2806 (mod 18 · 29 · 65) and therefore, letting m = 2 · 2806= 5612, we have N ′m ≡ 0 (mod 37 · 59 · 131). Let n ≥ 3 and εm(n) (m an even integer ≥ 2) be as in Section 3. If m ≡ l (mod ϕ(n)) for m, l ≥ 2, then we know that (εm(n), n) = d if and only if (εl(n), n) = d. If p − 1 - m for all prime divisors p of n, then Hm(n)≡ Hl(n) (mod n) by Corollary 4.3, hence putting n0 = n/d, we have εm(n) d βm ≡ εl(n) d βl (mod n0), where (εm(n)/d, n0) = (εl(n)/d, n0) = 1. Consequently, if m≡ l (mod ϕ(n)), then (N ′m, n0) = (N ′l , n0). Here we see p−1 - m for any prime divisor p of N ′m. Indeed, if p−1 | m, then Bm 6∈ Zp by Theorem 2.5, which is contrary to (N ′m, D′m) = 1. In particular, taking n = |N ′m| > 3 and d = (εm(n), N ′m), it is easily seen that if m≡ l (mod ϕ(N ′m)), then |N ′m/d|= (N ′ l , N ′m/d). The next theorem can be deduced immediately from Theorem 2.8, however we would like to give the proof from a different viewpoint without of use primes. Theorem 6.2. The set S = {|N ′2m| | m = 1, 2,3, ...} contains infinitely many elements that are relatively prime in pairs. Proof. Assume that |N ′2m1 |, |N ′2m2 |, ..., |N ′2mk | are all the elements in S which are relatively prime in pairs. Since |β2m| → ∞ as m→∞, we can choose x ≥ 1 such that ϕ(M1) ≥ 6 and |βϕ(M1)| > 1 for M1 = x ∏k i=1 |N ′ 2mi |. Then we see from Theorem 2.5 that ∏ p|M1 p divides D′ ϕ(M1) , hence (|N ′ ϕ(M1) |, M1) = 1. Next put M2 = M1|N ′ϕ(M1) |. Then 1 < |βϕ(M1)| < |βϕ(M2)| since 6 ≤ ϕ(M1) < ϕ(M2). By the same reason as mentioned above for M1, we see again (|N ′ ϕ(M2) |, M2) = 1. Repeating these procedures, we are able to produce an infinite sequence |N ′2m1 |, ..., |N ′2mk |, |N ′ϕ(M1) |, ..., |N ′ϕ(Mn) |, ..., which consists of relatively prime members in pairs. This is however contrary to the assump- tion and therefore the assertion follows. It is obvious that Theorem 6.2 contains the statement that there are infinitely many irreg- ular primes. Acknowledgment The author is grateful to the anonymous referee for his/her useful comments on the first version of this paper. REFERENCES 21 References [1] T. Agoh, On Bernoulli and Euler numbers, Manuscripta Math. 61 (1988), 1-10. [2] T. 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