On Congruences Involving Euler Polynomials and the Quotients of Fermat

J. Appl. Math. & Informatics Vol. 39(2021), No. 3 - 4, pp. 405 - 417 https://doi.org/10.14317/jami.2021.405

ON CONGRUENCES INVOLVING EULER POLYNOMIALS AND THE QUOTIENTS OF FERMAT

DOUK SOO JANG

Abstract. The aim of this paper is to provide the residues of Euler polynomials modulo p2 in terms of alternating sums of like powers of numbers in arithmetical progression. Also, we establish the analogue of a classical congruence of Lehmer.

AMS Mathematics Subject Classification : 11A07, 11B68. Key words and phrases : Euler polynomials, congruences, Fermat quotient.

1. Introduction

The associated Euler number Em(0) (m = 0, 1, 2,...) is defined by means of the following generating function ∑∞ 2 E (0) = m tm (1) et + 1 m! m=0 (see [1, 2, 8, 13]). Moreover, we can write ( ) ( ) −1 ∑∞ i 2 1 − et 1 − et = 1 − = , (2) et + 1 2 2 i=0 which converges for |1−et| < 2 and hence small values of t. The first few non-zero ones are: 1 1 1 17 E0(0) = 1,E1(0) = − ,E3(0) = ,E5(0) = − ,E7(0) = ,.... 2 4 2 8 2 − In particular, Em(0)’s are all rational numbers. Since et+1 1 is an odd function (i.e., if f(t) = f(t), so all of the signs are switched), we see that

Em(0) = 0 for m an even integer greater than or equal to 2. (3)

The Euler polynomials Em(t)(m = 0, 1, 2,...) is defined by ( ) ∑m m − E (t) = E (0)tm i, (4) m i i i=0 ( ) m n! where i = i!(m−i)! (see, for details, [13]). The associated Euler number numbers and Euler polynomials are classical and important in number theory and arise in some combinatorial contexts. For example, see E. Lehmer [9], Sun [13], and Wagstaff [15]. Equation (4) is actually the standard way to define andcom- pute the Euler polynomials inductively. The Euler polynomial Em(t) of degree m can be rewritten as the unique polynomial solution of the equation m Em(t + 1) + Em(t) = 2t , m ≥ 0 (5) (see [1, 2, 8, 13]). Also, Euler polynomials satisfy the identity m Em(1 − t) = (−1) Em(t), (6) which follows from (1) and (4). In particular, we have ( ) m 1 Em = 2 Em , (7) 2 where Em are the ordinary( Euler) numbers (see [12, 13, 15]), and as a result of (6), 1 ̸ we must have Em = Em 2 = 0 whenever m is odd. Therefore Em = Em(0), in fact [13, p. 374, (2.1)]

2 m+1 Em(0) = (1 − 2 )Bm+1, (8) m + 1 where Bm means the Bernoulli numbers and m ≥ 0. Following work of Friedmann and Tamarkin [5], E. Lehmer [9] considered Bernoulli numbers and polynomials modulo primes and prime powers, and showed many identities and combinatorial interpretations involving harmonic numbers. See E. Lehmer [9] for connections between Fermat quotients and Fermat’s last theorem. The motivation of the paper is to generalize the identities of E. Lehmer [9] on Euler polynomials. The results are presented in Section 2 and Section 3, and our proof is based on certain arithmetical identities and congruences for some alternating sums.

2. Results The following result concerning Euler polynomials was recently presented in [8, p. 2169, Lemma 2.1]. Lemma 2.1. Let m ≥ 0 be integers. Then ( ) ( ) ∑m k ∑k 1 k j m Em(t) = (−1) (j + t) . (9) 2 j i=0 j=0 On Congruences Involving Euler Polynomials and the Quotients of Fermat 407

In particular, if m ≥ 0 be integers, then m m 2 Em(t) ∈ Z[t] and 2 Em(0) ∈ Z. (10) This means that we have the following relationship between the ordinary Euler numbers E2n and the associated Euler numbers Ei(0) where i = 0, 1,..., 2n: ( ) ∑2n 2n E = E (0) ∈ Z. 2n i i i=0

Remark 2.1. We easily see that Em(0) will not contain primes p in the denom- inator when p > 2. Since Euler polynomials satisfy many properties that are similar to those that Bernoulli polynomials satisfy, we would expect a result similar to Kummer’s congruence for associated Euler numbers Em(0). We have the following result. Lemma 2.2. For integers m ≥ 1 and primes p ≥ 3, we have

Em+p−1(0) ≡ Em(0) (mod p). (11) Proof. When m and n are positive integers, it follows at once from (4) that

Em(n) ≡ Em(0) (mod n). (12) Let n be an odd integer with n ≥ 1. Moreover it is easy to see that n∑−1 ( ) l l+1 Em(0) + Em(n) = (−1) Em(l) − (−1) Em(l + 1) . (13) l=0 Using (5) and (13) we have n∑−1 l m Em(0) + Em(n) = 2 (−1) l . (14) l=0 For any odd prime p, by (12), (14) yields the congruence n∑−1 l m+p−1 Em+p−1(0) = (−1) l (mod n). (15) l=0 Consequently from Fermat’s little Theorem, i.e., lm+p−1 ≡ lm (mod p) for 0 ≤ l < p, for n = p, (15) imply Kummer’s congruence for Em(0):

Em+p−1(0) ≡ Em(0) (mod p). This completes the proof. Remark 2.2. Euler, on page 499 in [4], introduced Euler polynomials to eval- uate the alternating sum (14). Carlitz and Levine [2] have also investigated Kummer’s congruence for ordinary Euler numbers (7). See Wagstaff, Jr. [15, Theorem 4] for a simple proof of Kummer’s congruence for ordinary Euler numbers. 408 Douk Soo Jang

Theorem 2.3. For integers n ≥ 1 and primes p ≥ 3, we have p { } ∑[ n ] 2k ( ) r−1 2k n 2k [ p ]+1 s 3 (−1) (p − rn) ≡ p E2k−1 (0) + (−1) n E2k (mod p ) 2 n n r=1 (16) and

p ∑[ n ] 2k+1 { ( )} r−1 2k+1 n [ p ]+1 s 2 (−1) (p−rn) ≡ E2k+1 (0) + (−1) n E2k+1 (mod p ), 2 n r=1 (17) where s is the least positive residue of p (mod n) and k ≥ 1.

Proof. If in (5), alternating,[ ] adding and subtracting this identity with t = (p − p rn)/n, r = 1, 2,..., n , where [x] is the greatest integer not exceeding x, n and p are positive integers with n < p, for each case, gives the formula

p ( ) ( ) ( [ ]) ∑[ n ] − m p p +1 p p r−1 p rn E + (−1)[ n ] E − = 2 (−1) . m n m n n n r=1 This implies that

p ∑[ n ] m { ( ) ( )} r−1 m n p [ p ]+1 s (−1) (p − rn) = Em + (−1) n Em , (18) 2 n n r=1 where we have written s for the least positive residue of p modulo n. Setting m = 2k, k ≥ 1 and t = p/n, where p is an odd prime > n, in (4), by (3) and (10), we get the congruence ( ) ( ) ∑2k ( )2k−r ( ) p 2k p p 3 E = E (0) ≡ 2k E − (0) (mod p ). (19) 2k n r r n n 2k 1 r=0 Similarly we find for m = 2k + 1 with k ≥ 1 ( ) ( )2 p p 3 E ≡ E (0) + k(2k + 1)E − (0) (mod p ), (20) 2k+1 n 2k+1 2k 1 n since E2k(0) = 0 with k ≥ 1. Substituting these results into (18), we obtain the theorem.

Corollary 2.4. For integers n ≥ 1 and primes p ≥ 3, we have

p ∑[ n ] ( ( )) r−1 2k+1 1 [ p ]+1 s (−1) r ≡ − E2k+1(0) + (−1) n E2k+1 2 n (21) r=1 ( ) [ p ]+1 p s 2 + (−1) n (2k + 1)E2k (mod p ) 2n n On Congruences Involving Euler Polynomials and the Quotients of Fermat 409 and

p ( ) ∑[ n ] ( ) ( ) r−1 2k [ p ] 1 s p s 2 (−1) r ≡ (−1) n − E2k + kE2k−1 (mod p ), (22) 2 n n n r=1 where s is the least positive residue of p (mod n) and k ∈ N.

Proof. Since

p ∑[ n ] − (−1)r 1(p − rn)m r=1    p p  (23) ∑[ n ] ∑[ n ]  − pm − − ≡ (−1)mnm (−1)r 1rm − (−1)r 1rm 1 (mod p2),  n  r=1 r=1 congruences (16) and (17) may be combined to give sums of like powers of numbers less than [p/n] . From (23), we can write

p ∑[ n ] − nm (−1)r 1rm r=1    p p  ∑[ n ] ∑[ n ]  − − − ≡ (−1)m (−1)r 1(p − rn)m − pm (−1)r 1(p − rn)m 1 (mod p2),   r=1 r=1 (24) where m > 1. Now we put m = 2k − 1 with k > 1. Then, from (16), (17) and (24), we have

p ∑[ n ] ( ( )) r−1 2k+1 1 [ p ]+1 s (−1) r ≡ − E2k+1(0) + (−1) n E2k+1 2 n (25) r=1 ( ) [ p ]+1 p s 2 + (−1) n (2k + 1)E2k (mod p ). 2n n

Similarly, for m = 2k with k > 1, we obtain (22). This completes the proof.

Remark 2.3. Congruences (16) and (17) may be thought of as generalizations of Glaisher’s results [7] on Euler polynomials, while congruences (21) and (22) give generalizations of Vandiver’s results on Euler polynomials( ) whenever possible. s Both sets of formulas depend on the evaluation of Em n . 410 Douk Soo Jang ( ) s The values of Em n can be tabulated as follows:

E2k+1(1) = −E2k+1(0), k ≥ 0, ( ) ( ) ( ) 1 1 2k 1 E2k+1 − = − ,E2k+1 = 0, k ≥ 0, ( )2 2 ( ) ( 2 ) (26) 1 − 2 1 − 1 ≥ E2k+1 = E2k+1 = 1 2k+1 E2k+1(0), k 0, (3) (3) 2 3 1 3 E2k+1 = −E2k+1 , k ≥ 0. 4 4 ( ) s These evaluations of Em n are well known. Example 2.5. It follows readily from (17) and (26) that (p∑−1)/2 r−1 2k+1 2k 2 (−1) (p − 2r) ≡ 2 E2k+1(0) (mod p ), (27) r=1

p ∑[ 3 ] − (−1)r 1(p − 3r)2k+1 r=1 { ( ) 2k+1 E − E 1 p2 if p ≡ ≡ 3 2k+1(0) 2k+1 ( 3 ) (mod ) 1 (mod 3) (28) 2 2 2 E2k+1(0) + E2k+1 (mod p ) if p ≡ 2 (mod 3) ( (3 )) 2k+1 3 1 2 ≡ E2k+1(0) − E2k+1 (mod p ), p > 3, 2 3

p ∑[ 4 ] − (−1)r 1(p − 4r)2k+1 r=1 { p ( ) − − [ 4 ] 1 2 ≡ 2k E2k+1(0) ( 1) E2k+1 4 (mod p ) if p 1 (mod 4) ≡ 2 · 4 p ( ) 3 2 (29) E (0) − (−1)[ 4 ]E (mod p ) if p ≡ 3 (mod 4) ( 2k+1 2k+1 (4 )) 2k [ p ] 1 ≡ 2 · 4 E2k+1(0) ∓ (−1) 4 E2k+1 4 (mod p2) if p ≡ ±1 (mod 4), p > 5. ( ) s Example 2.6. Next we will give the results of substituting Em n in (21) and (22) for n = 1 and 2. If n = 1, (21) and (22) are of course the same as (−1)×(17) and (−1)×(16), respectively. We obtain easily ∑p r−1 2k+1 2 (−1) r ≡ −E2k+1(0) (mod p ), r=1 (30) ∑p r−1 2k 2 (−1) r ≡ −pkE2k−1(0) (mod p ). r=1 On Congruences Involving Euler Polynomials and the Quotients of Fermat 411

If n = 2, (21) becomes (p−1)/2 ( ( )) ∑ − r−1 2k+1 1 p 1 p 1 2 (−1) r ≡ − E2k+1(0) + (−1) 2 (2k + 1)E2k (mod p ), 2 2 2 r=1 (31) since E2k+1(1/2) = 0 with k ≥ 0. Similarly, (22) becomes (p∑−1)/2 ( ) r−1 2k p+1 1 1 2 (−1) r ≡ (−1) 2 E2k (mod p ), (32) 2 2 r=1 while (16) gives (p∑−1)/2 ( ( )) r−1 2k 2k−1 p+1 1 3 (−1) (p−2r) ≡ 2 pkE2k−1(0) + (−1) 2 E2k (mod p ), 2 r=1 (33) where k > 1. Hence, (33) reduces to the congruence − (p∑1)/2 − p+1 r−1 2k ( 1) 2 (−1) (p − 2r) ≡ E2k (mod p), (34) 2 r=1 where Ek is the kth ordinary Euler numbers (see (7)) and k > 1. 3. Applications Let p be an odd prime and a an integer not divisible by p. The quotient ap−1 − 1 q (a) = (35) p p is called the Fermat quotient of p with base a, which is an integer according to the Fermat Little Theorem. This quotient has been extensively studied because of its links to numerous question in number theory. A classical congruence, due to F.G. Eisenstein [3] in 1850, asserts that for a prime p ≥ 3, ∑p−1 1 r−1 1 qp(2) ≡ (−1) (mod p), (36) 2 r r=1 which was extended in 1861 by J.J. Sylvester [14] and in 1901 by Glaisher [6, pp. 21–22] as (p∑−1)/2 1 1 qp(2) ≡ − (mod p). (37) 2 r r=1 The above congruence was generalized in 1905 by M. Lerch in the first paper of substance on Fermat quotients [10] (see [11, pp. 949–950]). Theorem 3.1. Let t(p − 1) ̸≡ 2 (mod p − 1) and (p, t) = 1. Then ∑p−1 a 1 (−1) qp(a) ≡ − E − − (0) (mod p). (38) 2 t(p 1) 1 a=1 412 Douk Soo Jang

Proof. Observe that

(p∑−1)/2 ∑p−1 ∑p−1 − 2 (2r − 1)2k = r2k + (−1)r 1r2k. (39) r=1 r=1 r=1 As early as 1938, Lehmer [9, (15)] established the following interesting congruence ∑p−1 2k 2 r ≡ pB2k (mod p ) if 2k ̸≡ 2 (mod p − 1), (40) r=1 where Bk is the kth Bernoulli numbers (see [13]). From (30), (35) and (40), it follows that (p∑−1)/2 2k 1 2 (2r − 1) ≡ (pB2k − pkE2k−1(0)) (mod p ) (41) 2 r=1 if 2k ̸≡ 2 (mod p − 1). Further, in virtue of (35), we obtain that

∑p−1 ∑p−1 (p∑−1)/2 a a (−1) qp(a) = qp(a) − 2 (−1) qp(2a − 1). (42) a=1 a=1 a=1 Recall ([9, p. 354]) that

t(p−1) 2 a = 1 + ptqp(a) (mod p ), (43) where t ∈ N. Thus by (42) and (43), we have ∑p−1 a pt (−1) qp(a) a=1 ∑p−1 ( ) (p∑−1)/2 ( ) − − ≡ at(p 1) − 1 − 2 (2a − 1)t(p 1) − 1 (mod p2) a=1 a=1 ∑p−1 (p∑−1)/2 (44) − − ≡ at(p 1) − 2 (2a − 1)t(p 1) (mod p2) a=1 a=1 − p(p 1)t 2 ≡ E − − (0) (mod p ) if t(p − 1) ̸≡ 2 (mod p − 1) 2 t(p 1) 1 pt 2 ≡ − E − − (0) (mod p ) if t(p − 1) ̸≡ 2 (mod p − 1), 2 t(p 1) 1 that is, ∑p−1 a 1 (−1) qp(a) ≡ − E − − (0) (mod p) (45) 2 t(p 1) 1 a=1 if t(p − 1) ̸≡ 2 (mod p − 1) and (p, t) = 1, and the proof is completed. On Congruences Involving Euler Polynomials and the Quotients of Fermat 413

Theorem 3.2. For integers k ≥ 1 and primes p ≥ 3, we have ∑p−1 − 1 (−1)r 1r2k+1q (r) ≡ (E (0) − E (0)) (mod p) (46) p p 2k+1 2k+p r=1 and ∑p−1 − r−1 2k p 1 (−1) r qp(r) ≡ − E2k−1(0) (mod p). (47) 2 r=1 Proof. Now we in a position to transform the above sums into sums involving Fermat’s quotients by means of the relation a−1 m+p−1 m a−1 m (−1) (a − a ) = (−1) pa qp(a). (48) Hence, (48) may be written as ∑p−1 ∑p−1 ∑p−1 r−1 m r−1 m+p−1 r−1 m p (−1) r qp(r) = (−1) r − (−1) r . (49) r=1 r=1 r=1 Putting m = 2k + 1 in (49), from (30), we find ∑p−1 ∑p−1 ∑p−1 r−1 2k+1 r−1 2k+p r−1 2k+1 p (−1) r qp(r) = (−1) r − (−1) r r=1 r=1 r=1 (50) 2 ≡ E2k+1(0) − E2k+p(0) (mod p ), that is, ∑p−1 − 1 (−1)r 1r2k+1q (r) ≡ (E (0) − E (0)) (mod p), (51) p p 2k+1 2k+p r=1 where k ∈ N. Putting m = 2k in (49), from (30), we find ∑p−1 ∑p−1 ∑p−1 r−1 2k r−1 2k+p−1 r−1 2k p (−1) r qp(r) = (−1) r − (−1) r r=1 r=1( r=1 ) − 2k + p 1 2 ≡ p kE2k−1(0) − E2k−1+p−1(0) (mod p ) ( 2 ) − 2k + p 1 2 ≡ p kE2k−1(0) − E2k−1(0) (mod p ) 2 (by (11)) ( ) − 2k + p 1 2 ≡ p k − E2k−1(0) (mod p ), 2 (52) that is, ∑p−1 − r−1 2k p 1 (−1) r qp(r) ≡ − E2k−1(0) (mod p), (53) 2 r=1 where k ∈ N. This completes the proof. 414 Douk Soo Jang

Theorem 3.3. Let n > 1, α > 1 and p > 3. Then  ( )  s [ p ] E α − − (−1) n E α − (0)  ϕ(p ) 1 n ϕ(p ) 1 [ ] p  2 p ∑[ n ] ϕ(pα)−1  (mod p ) if n is odd − r−1 1 ≡ n ( 1)  p − nr 2  ( ) p r=1  s −E α − − [ n ]E α −  ϕ(p ) 1 n + ( 1) ϕ(p ) 1[(0)]  2 p (mod p ) if n is even. (54) Proof. Also α ∈ N and n > 1. If, by a slight change in notation, we set  p [∑n ] ( ) [ ]  r−1 s m p p  (−1) + r − 1 if is odd [ ]  n n ∑n ( )m r=1 − p (−1)r 1 − r = n  r=1  p [∑n ] ( ) [ ]  − r s − m p ( 1) n + r 1 if n is even r=1 (55) p −1 [ n∑] ( ) [ ]  − r s m p  ( 1) n + r if n is odd  r=0 =   p −1 [ n∑] ( ) [ ]  − r−1 s m p ( 1) n + r if n is even, r=0 then we have

p ∑[ n ] − 1 (−1)r 1 p − nr r=1 p ∑[ n ] − α − ≡ (−1)r 1(p − nr)ϕ(p ) 1 (mod pα) r=1 [ p ] ∑n ( )ϕ(pα)−1 (56) α − − p ≡ nϕ(p ) 1 (−1)r 1 − r (mod pα) n r=1 p − [ n ] 1  ∑ ( )ϕ(pα)−1 [ ]  (−1)r s + r (mod pα) if p is odd α − n n = nϕ(p ) 1 r=0  p −1 [ n∑] ( ) α − [ ]  − r−1 s ϕ(p ) 1 α p ( 1) n + r (mod p ) if n is even, r=0 where ϕ(n) denotes Euler’s totient function. It is well known that (cf. [13]) n∑−1 r m 1 n−1 (−1) (t + r) = (Em(t) + (−1) Em(t + n)). (57) 2 r=0 On Congruences Involving Euler Polynomials and the Quotients of Fermat 415

From (56) and (57), it follows that  ( ) ( [ ])  s [ p ]−1 s p E α − + (−1) n E α − +  ϕ(p ) 1 n ϕ(p ) [1 ]n n p  α p ∑[ n ] ϕ(pα)−1  (mod p ) if n is odd − r−1 1 ≡ n ( 1)  p − nr 2  ( ) p ( [ ]) r=1  s s p −E α − − [ n ]E α −  ϕ(p ) 1 n + ( 1) ϕ(p ) 1[ n] + n  α p (mod p ) if n is even. (58) Since Eϕ(pα)−2(0) = 0 with α > 1 and p > 3, we have ( [ ]) ( ) s p p E α − + = E α − ϕ(p ) 1 n n ϕ(p ) 1 n p 2 ≡ E α − (0) + E α − (0) (mod p ) ϕ(p ) 1 n ϕ(p ) 2 2 ≡ Eϕ(pα)−1(0) (mod p ), which has the paraphrase  ( )  s [ p ] E α − − (−1) n E α − (0)  ϕ(p ) 1 n ϕ(p ) 1 [ ] p  2 p ∑[ n ] ϕ(pα)−1  (mod p ) if n is odd − r−1 1 ≡ n ( 1)  p − nr 2  ( ) p r=1  s −E α − − [ n ]E α −  ϕ(p ) 1 n + ( 1) ϕ(p ) 1[(0)]  2 p (mod p ) if n is even, (59) where α > 1 and p > 3. This completes the proof. Example 3.4. (54) implies that, for n = 2 and 3, (p∑−1)/2 r−1 1 ϕ(pα)−2 2 (−1) ≡ 2 E α − (0) (mod p ), (60) p − 2r ϕ(p ) 1 r=1 p ( ( )) ∑[ 3 ] ϕ(pα)−1 r−1 1 3 1 2 (−1) ≡ E α − (0) − E α − (mod p ), p − 3r 2 ϕ(p ) 1 ϕ(p ) 1 3 r=1 (61) where we use the fact that for p > 3, [ ] p p ≡ 1 (mod 3) ⇔ ∃ even integer k > 1 satisfying p = 3k + 1 ⇔ is even, [ 3] p p ≡ 2 (mod 3) ⇔ ∃ odd integer k ≥ 1 satisfying p = 3k + 2 ⇔ is odd 3 m and Em(1 − t) = (−1) Em(t). Example 3.5. By Fermat’s quotients (35), we obtain

1 − q (a) = ap 2 − p p , a a 416 Douk Soo Jang so that, after a is replaced by p − nr,

p p p ∑[ n ] ∑[ n ] ∑[ n ] − 1 − − − 1 p (−1)r 1 q (p − nr) = (−1)r 1(p − nr)p 2 − (−1)r 1 . p − nr p p − nr r=1 r=1 r=1 (62) Now, if we write n = 2 in (17), then (p∑−1)/2 r−1 p−2 p−3 2 (−1) (p − 2r) ≡ 2 Ep−2(0) (mod p ), (63) r=1 where we have used E2k+1(1/2) = 0 with k ≥ 0. Next, we put n = 2 in (62) and use (60), (63), we get (p∑−1)/2 r−1 1 (−1) qp(p − 2r) p − 2r r=1 ( ) (64) 1 p−2 ϕ(pα)−1 ≡ 2 Ep−2(0) − 2 E α − (0) (mod p), 2p ϕ(p ) 1 where α > 1 and p > 3. Simiarly, the evaluation (62) with n = 3 provides a new expression for p ∑[ 3 ] r−1 1 (−1) qp(p − 3r) p − 3r r=1 by using (28) and (61).

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Douk Soo Jang received Ph.D. degree from Kyungnam University. His main research area is analytic number theory. Recently, his main interests focus on zeta and multiple zeta functions, Bernoulli and Euler numbers and polynomials. Division of Mathematics, Science, and Computers, Kyungnam University, 7(Woryeong- dong) kyungnamdaehak-ro, Masanhappo-gu, Changwon-si, Gyeongsangnam-do 51767, Ko- rea. e-mail: [email protected]