Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2
aa r X i v : . [ m a t h . N T ] N ov REDUCING THE ERD ˝OS-MOSER EQUATION1 n + 2 n + · · · + k n = ( k + 1) n MODULO k AND k JONATHAN SONDOW AND KIEREN MACMILLAN
Abstract.
An open conjecture of Erd˝os and Moser is that the only solution of the Diophantineequation in the title is the trivial solution 1+2 = 3. Reducing the equation modulo k and k , we givenecessary and sufficient conditions on solutions to the resulting congruence and supercongruence.A corollary is a new proof of Moser’s result that the conjecture is true for odd exponents n . We alsoconnect solutions k of the congruence to primary pseudoperfect numbers and to a result of Zagier.The proofs use divisibility properties of power sums as well as Lerch’s relation between Fermat andWilson quotients. Introduction
Around , Erd˝os and Moser studied the Diophantine equation(1.1) 1 n + 2 n + · · · + k n = ( k + 1) n and made the following conjecture. Conjecture 1.
The only solution of (1.1) in positive integers is the trivial solution . Moser [13] proved the statement when n is odd or k < . In Schinzel showed that in anysolution, k is even [12, p. 800]. An extension of Schinzel’s theorem to a generalization of equation(1.1) was given in by Moree [10, Proposition 9]. For a recent elementary proof of a specialcase, see MacMillan and Sondow [9, Proof of Proposition 2].Many other results on the Erd˝os-Moser equation (1.1) are known, but it has not even beenestablished that there are only finitely many solutions. For surveys of work on this and relatedproblems, see Butske, Jaje, and Mayernik [1], Guy [4, D7], and Moree [11].In the present paper, we first approximate equation (1.1) by the congruence(1.2) 1 n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) . In Section 2, we give necessary and sufficient conditions on n and k (Theorem 1), and we show thatif a solution k factors into a product of 1 , ,
3, or 4 primes, then k = 2 , ,
42, or 1806, respectively(Proposition 1). In Section 3, we relate solutions k to primary pseudoperfect numbers and to aresult of Zagier. In the final section, Theorem 1 is extended to the supercongruence (Theorem 2)1 n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) . Our methods involve divisibility properties of power sums, as well as Lerch’s formula relatingFermat and Wilson quotients.
Mathematics Subject Classification.
Primary 11D61; Secondary 11D79, 11A41.
Key words and phrases.
Chinese Remainder Theorem, congruence, Diophantine equation, Egyptian fraction,Erd˝os-Moser equation, Fermat quotient, perfect, power sum, primary pseudoperfect number, prime, pseudoperfect,square-free, supercongruence, Wilson quotient.
As applications of Theorems 1 and 2, we reprove Moser’s result that Conjecture 1 is true for oddexponents n (Corollary 1), and for even n we show that a solution k to (1.1) cannot be a primarypseudoperfect number with 8 or fewer prime factors (Corollary 5). In a paper in preparation, wewill give other applications of our results to the Erd˝os-Moser equation.2. The Congruence n + 2 n + · · · + k n ≡ ( k + 1) n (mod k )We will need a classical lemma on power sums. (An empty sum will represent 0, as usual.) Definition 1.
For integers n ≥ a ≥
1, let Σ n ( a ) denote the power sum Σ n ( a ) := 1 n + 2 n + · · · + a n . Set Σ n (0) := 0. Lemma 1. If n is a positive integer and p is a prime, then Σ n ( p ) ≡ ( − p ) , ( p − | n, p ) , ( p − ∤ n. Proof.
Hardy and Wright’s proof [5, Theorem 119] uses primitive roots. For a very elementaryproof, see MacMillan and Sondow [8]. (cid:3)
We now give necessary and sufficient conditions on solutions to (1.2).
Theorem 1.
Given positive integers n and k , the congruence (2.1) 1 n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) holds if and only if prime p | k implies (i). n ≡ p − , and (ii). kp + 1 ≡ p ) .In that case k is square-free, and if n is odd, then k = 1 or .Proof. First note that if n, k, p are any positive integers with p | k , then(2.2) Σ n ( k ) = k/p X h =1 p X j =1 (( h − p + j ) n ≡ kp Σ n ( p ) (mod p ) . Now assume that (i) and (ii) hold whenever prime p | k . Then, using Lemma 1, both Σ n ( p )and k/p are congruent to − p , and so Σ n ( k ) ≡ p ), by (2.2). Thus, as (ii) implies k is square-free, k is a product of distinct primes each of which divides Σ n ( k ) −
1. It follows thatΣ n ( k ) ≡ k ), implying (2.1).Conversely, assume that (2.1) holds, so that Σ n ( k ) ≡ k ). If prime p | k , then (2.2) gives kp Σ n ( p ) ≡ p ), and so Σ n ( p ) p ). Now Lemma 1 yields both ( p − | n , proving (i),and Σ n ( p ) ≡ − p ), implying (ii).If n is odd, then by (i) no odd prime divides k . As k is square-free, k = 1 or 2. (cid:3) Corollary 1.
The only solution of the Erd˝os-Moser equation with odd exponent n is .Proof. Given a solution with n odd, Theorem 1 implies k = 1 or 2. But k = 1 is clearly impossible,and k = 2 evidently forces n = 1. (cid:3) EDUCING THE EQUATION 1 n + 2 n + · · · + k n = ( k + 1) n MODULO k AND k Recall that, when x and y are real numbers, x ≡ y (mod 1) means that x − y is an integer. Corollary 2.
A given positive integer k satisfies the congruence (2.1) , for some exponent n , if andonly if the Egyptian fraction congruence (2.3) 1 k + X p | k p ≡ holds, where the summation is over all primes p dividing k . In that case, k is square-free, and n isany number divisible by the least common multiple LCM { p − prime p | k } .Proof. Condition (2.3) is equivalent to the congruence(2.4) 1 + X p | k kp ≡ k ) , which in turn is equivalent to condition (ii) in Theorem 1, since each implies k is square-free. Thetheorem now implies the corollary. (cid:3) Remark 1.
In (2.3) we write ≡ ≡ k with at most four(distinct) prime factors. First we prove a lemma. (An empty product will represent 1, as usual.) Lemma 2.
Let k = p p · · · p r , where the p i are primes. If k satisfies the integrality condition (2.3) ,then for any subset S ⊂ { p , . . . , p r } , there exists an integer q = q ( r, S ) such that (2.5) q Y p ∈ S p = 1 + X p ∈ S kp . Proof.
This follows from Corollary 2 and Theorem 1 condition (ii), using the Chinese RemainderTheorem. For an alternate proof, denote the summation in (2.5) by Σ, and note that (2.3) implies1 + Σ ≡ kp ≡ p ), for each p ∈ S . Then, since k is square-free, Q p ∈ S p divides 1 + Σ, andthe lemma follows. (cid:3) Proposition 1.
Let k be a product of r primes. Suppose n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) , forsome exponent n > ; equivalently, suppose (2.3) holds. If r = 0 , , , , , then k = 1 , , , , ,respectively.Proof. Theorem 1 implies k = p p · · · p r , where p < p < · · · < p r are primes.( r = 0 , r = 0, then k = 1. If r = 1, then k = p is prime, and (2.4) yields k |
2, so that k = 2.( r = 2). For k = p p , congruence (2.4) gives p | ( p + 1). Since p ≥ p + 1, it follows that p = p + 1. As p and p are prime, p = 3 and p = 2, and hence k = 6.( r = 3 , k = p · · · p r , where p < · · · < p r are primes, then by Lemma 2, for i = 1 , . . . , r there exists an integer q i such that q i p i = kp i + 1. In particular,(2.6) q r p r = p · · · p r − p r − + 1 . Hence if r >
2, so that p r − < p r −
1, then q r < p · · · p r − . We also have q r q r − p r − = q r ( p · · · p r − p r + 1) = p · · · p r − ( p · · · p r − + 1) + q r , JONATHAN SONDOW AND KIEREN MACMILLAN and so(2.7) p r − < p r − | ( p · · · p r − + q r ) < p · · · p r − . Now take r = 3, so that k = p p p . Then q p = p p + 1 and q p = p p + 1, for some integers q and q . By (2.7) we have p < p | ( p + q ) < p . Hence p = p + q . Substituting q = p − p into q p = p p + 1 yields p | ( p p − p | ( p p + 1), we conclude that p |
2. Therefore p = 2. Then p | (2 p + 1). As p > p , we get p = 2 p + 1. Then we have p | (2 p + 1) = 4 p + 3,and so p |
3. Therefore p = 3, and hence p = 7. Thus k = 2 · · r = 4. Lemma 2 with S = { p , . . . , p r } and r = 4 gives an integer q such that qp p p p = p ( p p + p p + p p ) + p p p + 1 < p p p , so that qp <
4. Hence q = 1, and p = 2 or 3. Now( p − p p p = p ( p p + p p + p p ) + 1 < p p p , and so ( p − p < p . This implies p = 2 and p = 3. Thus k = 2 · p p . Then (2.6) and (2.7)give q p = 6 p + 1 and 3 < p | (6 + q ) <
12. The only solution is ( q , p , p ) = (1 , , k = 2 · · ·
43 = 1806. This completes the proof. (cid:3)
Example 1.
Take k = 1806 in Corollary 2. Since 1806 = 2 · · ·
43 and11806 + 12 + 13 + 17 + 143 = 1 , and since LCM(1 , , ,
42) = 42, one solution of (2.1) is1 + 2 + · · · + 1806 ≡ (mod 1806) . Primary Pseudoperfect Numbers
Recall that a positive integer is called perfect if it is the sum of all its proper divisors, and pseudoperfect if it is the sum of some of its proper divisors [4, B2].
Definition 2. (From [1].) A primary pseudoperfect number is an integer
K > K + X p | K p = 1 , where the summation is over all primes dividing K .Multiplying (3.1) by K , we see that K is square-free, and that every primary pseudoperfectnumber, except 2, is pseudoperfect. Corollary 3.
Every primary pseudoperfect number K is a solution to the congruence (2.1) , forsome exponent n .Proof. This is immediate from Definition 2 and Corollary 2. (cid:3)
A priori, the equality (3.1) is a stronger condition than the congruence (2.3) in Corollary 2.However, (3.1) and (2.3) may in fact be equivalent, because all the known solutions of (2.3) alsosatisfy (3.1) — see [1]. In other words, primary pseudoperfect numbers may be the only solutions k to the congruence (2.1), except for k = 1.According to [1], Table 1 contains all primary pseudoperfect numbers K with r ≤ r = 1 , , . . . ,
8, there exists exactly one such K (as conjectured EDUCING THE EQUATION 1 n + 2 n + · · · + k n = ( k + 1) n MODULO k AND k by Ke and Sun [6] and by Cao, Liu, and Zhang [2]). No K with r > Table 1. (from [1]) The primary pseudoperfect numbers K with r ≤ r K prime factors1 2 22 6 2 ,
33 42 2 , ,
74 1806 2 , , ,
435 47058 2 , , , ,
316 2214502422 2 , , , , , , , , , , , , , , , , , , r = 1 , , , Table 2.
The known solutions to 1 n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) k n is any multiple of1 12 16 242 61806 4247058 3302214502422 23529052495396602 3108008490421583559688410706771261086 1863851053628494074457830Table 2 was calculated from Table 1, using Corollaries 2 and 3. Example 2.
The simplest case of the congruence (2.1) in which k has 8 prime factors is X j =1 j ≡ (mod 8490421583559688410706771261086) . Zagier gave three characterizations of the numbers 1 , , , , Proposition 2.
Each of the following five conditions is equivalent to k ∈ { , , , , } . (i) The congruence a k +1 ≡ a (mod k ) holds, for all a . (ii) k = p p · · · p r , where r ≥ , the p i are distinct primes, and ( p i − | k . (iii) k = p p · · · p r , where r ≥ and p i = p · · · p i − + 1 is prime, i = 1 , . . . , r . (iv) k is a product of at most primes, and n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) , for someexponent n . (v) Either k = 1 or k is a primary pseudoperfect number with or fewer prime factors. JONATHAN SONDOW AND KIEREN MACMILLAN
Proof.
For (i), (ii), (iii), see the solution to the first problem of Zagier [14]. Proposition 1 yields (iv).Corollary 3 and (iv) give (v). (cid:3) Supercongruences
If the conditions in Theorem 1 are satisfied, the following corollary shows that the congru-ence (2.2) can be replaced with a “supercongruence” (compare Zudilin [15]).
Corollary 4. If n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) and prime p | k , then (4.1) Σ n ( k ) ≡ kp Σ n ( p ) (mod p ) . Proof.
By Theorem 1, it suffices to prove the more general statement that, if prime p | k and( p − | n , and if either k = 2 or n is even, then (4.1) holds. Set a = k/p in the equation (2.2).Expanding and summing, we see thatΣ n ( k ) ≡ a Σ n ( p ) + 12 a ( a − np Σ n − ( p ) (mod p ) . If p >
2, then ( p − | n implies ( p − ∤ ( n − p | Σ n − ( p ). In case p = 2,either a = k/ | n , and each implies 2 | (1 / a ( a − n . In all cases, (4.1) follows. (cid:3) For an extension of Theorem 1 itself to a supercongruence, we need a definition and a lemma.
Definition 3.
By Fermat’s and Wilson’s theorems, for any prime p the Fermat quotient q p ( j ) := j p − − p , p ∤ j, (4.2)and the Wilson quotient W p := ( p − p are integers. Lemma 3 (Lerch [7]) . If p is an odd prime, then the Fermat and Wilson quotients are related by Lerch’s formula p − X j =1 q p ( j ) ≡ W p (mod p ) . Proof.
Given a and b with p ∤ ab , set j = ab in (4.2). Substituting a p − = pq p ( a ) + 1 and b p − = pq p ( b ) + 1, we deduce Eisenstein’s relation [3] q p ( ab ) ≡ q p ( a ) + q p ( b ) (mod p ) , which implies q p (( p − ≡ p − X j =1 q p ( j ) (mod p ) . On the other hand, setting j = ( p − pW p − p − ≥ q p (( p − ≡ W p (mod p ). This proves the lemma. (cid:3) EDUCING THE EQUATION 1 n + 2 n + · · · + k n = ( k + 1) n MODULO k AND k Theorem 2.
For n = 1 , the supercongruence (4.3) 1 n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) holds if and only if k = 1 or . For n ≥ odd, (4.3) holds if and only if k = 1 . Finally, for n ≥ even, (4.3) holds if and only if prime p | k implies (i). n ≡ p − , and (ii). kp + 1 ≡ p ( n ( W p + 1) −
1) (mod p ) . Proof.
To prove the first two statements, use Theorem 1 together with the fact that the congruences1 n + 2 n ≡ n ≡ ( − n ≡ − n ≥ n ≥ p denote a prime. By Theorem 1, we may assume that (i) holdsif p | k , and that k is square-free. It follows that the supercongruence (4.3) is equivalent to thesystem Σ n ( k ) ≡ ( k + 1) n (mod p ) , p | k. Corollary 4 and expansion of ( k + 1) n allow us to write the system as kp Σ n ( p ) ≡ nk (mod p ) , p | k. Since n is at least 2 and ( p − | n , we haveΣ n ( p ) ≡ Σ n ( p −
1) (mod p )= p − X j =1 ( j p − ) n/ ( p − . Substituting j p − = 1 + pq p ( j ) and expanding, the result isΣ n ( p ) ≡ p − X j =1 (cid:18) np − pq p ( j ) (cid:19) ≡ p − − np p − X j =1 q p ( j ) (mod p ) , (4.4)since n/ ( p − ≡ − n (mod p ). Now Lerch’s formula (if p is odd), together with the equality q (1) = 0 and the evenness of n (if p = 2), yieldΣ n ( p ) ≡ p − − npW p (mod p ) . Summarizing, the supercongruence (4.3) is equivalent to the system kp ( p − − npW p ) ≡ nk (mod p ) , p | k. It in turn can be written as kp + 1 ≡ − k (cid:0) n ( W p + 1) − (cid:1) (mod p ) , p | k. (4.5)On the right-hand side, we substitute k ≡ − p (mod p ) (deduced from (4.5) multiplied by p ), andarrive at (ii). This completes the proof. (cid:3) JONATHAN SONDOW AND KIEREN MACMILLAN
Corollary 5.
Let n ≥ be even and let K be a primary pseudoperfect number with r ≤ primefactors. (i). Then ( n, K ) is a solution of the supercongruence (4.3) if and only if either K = 2 , or K = 42 and n ≡
12 (mod 42) . (ii). The supercongruence (4.6) 1 n + 2 n + · · · + K n ≡ ( K + 1) n (mod K ) holds if and only if K = 2 and n ≥ . (iii). The Erd˝os-Moser equation has no solution ( n, k ) with k = K .Proof. (i). We use Table 1.( r = 1). Theorem 2 with k = p = 2 implies ( n,
2) is a solution to (4.3). (This can also be seendirectly from (4.3): both sides are congruent to 1 modulo 4.)( r = 2). Suppose k = 2 · | n , condition (ii) in Theorem 2 with p = 2 gives 3 + 1 = kp + 1 ≡ − k = 6.( r = 3). For k = 2 · ·
7, condition (i) in Theorem 2 requires 6 | n . Then (ii) is satisfied for p = 2and 3. For p = 7, we need 6 + 1 ≡ n −
1) (mod 49), which reduces to 3 n ≡ | n , only n ≡
12 (mod 42) gives a solution with k = 42.( r = 4). If k = 2 · · ·
43, condition (ii) with p = 2 rules out any solution.( r = 5). For k = 2 · · · ·
31, condition (i) gives 3 | n . As k ≡
6≡ − r = 6 , , K in Table 1 with r = 6 , , kp + 1 ≡ − p (mod p ), for p = 2 , ,
2, respectively. But the requirement is violated in eachcase, and so no solution exists.(ii). Part (i) implies that the only possible solutions ( n, K ) of (4.6) are K = 2, and K = 42 with n ≡
12 (mod 42).It is easy to check that ( n, K ) = (2 ,
2) is not a solution. To see that ( n,
2) is a solution when n ≥ n ≡ n (mod 2 ). Since 1 + 2 n = 1 + 4 n/ ≡ n = 9 n/ = (1 + 8) n/ ≡ K = 2 is proved.Now suppose (4.6) holds with K = 42 and n ≡
12 (mod 42). Since( K + 1) n ≡ nK + 12 n ( n − K (mod K ) , by setting n = 6 n we infer thatΣ n (42) ≡ − n + 31752 n ≡ ) . (4.7)But as n ≥ n (42) is congruent to 1 or 0 modulo 8according as the term is odd or even, and so Σ n (42) ≡
21 (mod 8). This contradicts (4.7), prov-ing (ii).(iii). This follows from (ii) and the fact that if k = 2 in the Erd˝os-Moser equation, then evidently n = 1. (cid:3) EDUCING THE EQUATION 1 n + 2 n + · · · + k n = ( k + 1) n MODULO k AND k Example 3.
The simplest cases of (i) are 1 + 2 ≡ (mod 2 ) and1 + 2 + · · · + 42 ≡ (mod 42 ) . An example of (ii) is 1 + 2 ≡ (mod 2 ). (More generally, one can show that1 n + 2 n ≡ n (mod 2 d ) , if 2 d − | n, for any positive integers n and d .)In light of Theorem 1 and Corollary 4, it is natural to ask whether Theorem 2 has an analogouscorollary about supercongruences modulo p . Conjecture 2. If n + 2 n + · · · + k n ≡ ( k + 1) n (mod k ) and prime p | k , then Σ n ( k ) ≡ kp Σ n ( p ) (mod p ) . Example 4.
For p = 2 , ,
7, one can compute that1 + 2 + · · · + 42 ≡ p (cid:0) + 2 + · · · + p (cid:1) (mod p ) . In fact, for p = 2 , , n (42) ≡ p Σ n ( p ) (mod p ) holds true not only when n ≡
12 (mod 42), but indeed for all n ≡ p = 7 (but notfor p = 2 or 3), apparently 6 | n implies p | Σ n − ( p ). (Compare p | Σ n − ( p ) in the proof ofCorollary 4.) Acknowledgments
We are very grateful to Wadim Zudilin for contributing Theorem 2 and some of the other resultsin Section 4. The first author thanks both the Max Planck Institute for Mathematics for itshospitality during his visit in October 2008 when part of this work was done, and Pieter Moreefor reprints and discussions of his articles on the Erd˝os-Moser equation. The second author thanksAngus MacMillan and Dr. Stanley K. Johannesen for supplying copies of hard-to-locate papers,and Drs. Jurij and Daria Darewych for underwriting part of the research.
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