1 x i. i=1 EVEN NUMBERS RAFAEL ARCE-NAZARIO, FRANCIS N. CASTRO, AND RAÚL FIGUEROA

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1 Volume, Number 2, Pages ISSN ON THE EQUATION n i= = IN DISTINCT ODD OR EVEN NUMBERS RAFAEL ARCE-NAZARIO, FRANCIS N. CASTRO, AND RAÚL FIGUEROA Abstract. In this paper we combine theoretical results and computer search to obtain information about the solutions of the equation n =. i= We calculate (a) m o(n) = min{max{ i n}} and (b) m e(n) = min{max{ i n}}, where the minimum is taken over all sets {} satisfying the above equation in distinct odd integers when 3 n 4 (for case a) and in distinct even integers when 3 n 29 (for case b). We compute the number of solutions of the above equation for: n = 3, 5 when {3 α 5 β 7 γ }; n 7 when {3 α 5 β γ }; n 23 when {3 α 5 β 3 γ }. We also compute max for {} satisfying the mentioned equation in distinct even integers. Finally, we compute lim inf(x n/x n k ) for fixed k.. Introduction In the book Unsolved Problems in Number Theory [7], Section D discusses the topic of Egyptian fractions. These are fractions that can be expressed as a finite sum n i= / of reciprocals of distinct positive integers. Special attention has been given to the equation (.) n i= =, where x < < x n. The number of solutions of Equation (.) for n 8 over distinct positive integers are given in [3]. Furthermore, in [4,, ], the authors computed the number of solutions of Equation (.) for n = 9, over distinct odd positive integers. A question proposed by Erdős and Graham about these fractions that appears in [7] is to determine the value of m(n), the min max, where the minimum is taken over all sets { } satisfying Equation (.). Section D Received by the editors February 26, 205, and in revised form February 4, Mathematics Subject Classification. Primary: D68; Secondary: Y50. Key words and phrases. Unit fractions, Egyptian fractions. 63 c 207 University of Calgary

2 64 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA in [7] contains a table showing m(k) for 3 n 28. It is observed that m(n) is nondecreasing for these values. In [2, 5], the authors proved that any rational number with an odd denominator can be written as the sum of distinct odd unit fractions. In [0], G. Martin computed the asymptotic behavior of m(n) for Equation (.). More precisely, he proved (.2) m(n) = min max i n = n e + O ( n log log 3n log 3n In this paper we consider the following problem: to determine the values of and m o (n) = min max i n m e (n) = min max i n, where are distinct odd(even) positive integers satisfying Equation (.). In fact, for each odd n, 3 n 4, we found the value of m o (n) and all the sequences where m o (n) is the minimum value for the corresponding n. The minimum n for which m o (n) exists is n = 9; the values m o (9) = 23 and m o () = 05, were independently found by P. Shiu [] and N. Burshtein [3]. In [], Arce et. al. computed m o (3) = 5. In this paper we compute m o (n) for 5 n 4. For the case m e (n) we compute m e (n) for 3 n 29. Our results imply that m o (n) and m e (n) are nondecreasing for n 4, 3 n 29, respectively. Note that Martin s result in [0] implies that n m e (n) = + (error term). e 2 In [5], Chen-Elsholtz-Jiang proved a criterion for existence of solutions of Equation (.) over subsets of the type S(p,..., p r ) = {p α pαr r α,..., α r N 0 }, where N 0 are the nonnegative integers. Let N n (p,..., p r ) be the set of the solutions of Equation (.) with S(p,..., p r ) where any solution contains at least one x j that is divisible by p i for i =,..., r. Note that in general N n (p,..., p r ) is not equal to T n (p,..., p r ) of [5], as N n (p,..., p r ) T n (p,..., p r ) and N n (p, p 2, p 3 ) = T n (p, p 2, p 3 ). It is not difficult to prove that N n (p, p 2 ) = 0, where p, p 2 are odd primes; see [5] for details. Hence the simplest case is N n (p, p 2, p 3 ). In [3], Burshtein proved that N (3, 5, 7) = 7. Using the information in [], we have N (p, p 2, p 3 ) = 0 except in the case of p = 3, p 2 = 5, and p 3 = 7. Motivated by the results of [5, 3, ], we compute N 3 (3, 5, 7) = 2034 and N 5 (3, 5, 7) = N n (3, 5, ) = 0 for n 5 and N 7 (3, 5, ) =. N n (3, 5, 3) = 0 for n 2 and N 23 (3, 5, 3) = 63. ).

3 EGYPTIAN FRACTIONS 65 We also prove N n (p,..., p r ) over S(p, p 2,..., p r ) for n n o odd and some p,..., p r odd primes. In this paper we study solutions of Equation (.) satisfying some restrictions. We prove the existence of solutions of Equation (.) where the last k terms are close to each other in a multiplicative sense. The optimization problem of finding m(n) was solved in [8, 4]. In this paper we solve the optimization problem of finding max{ } over positive even numbers of the Equation (.). The results in Section 2 are mainly computational, but in order to significantly reduce the computational time we had to develop some elementary constraints. We used some divisibility properties to obtain these constraints and they appear as theorem and corollaries in Section 2. The Tables in Section 2 summarize our results for m o (n), m e (n). A complete list of sequences can be found in our page rarce/dmath.html. In Section 3, we present the calculation of m o (27) and m o (29) to show that in some cases the constraints reduce the solution space enough that they can be completed by hand. In Section 4, we study the solutions of Equation (.) over S(p,..., p r ). In section 4 we also combine computational results with an identity to prove the solvability of Equation (.). In Section 5, we construct solutions of Equation (.) satisfying certain properties. Our results of this section imply that lim inf x n /x n k = for fixed k. In the last section, we answer the optimization problem for max{ } over positive even numbers. Our calculations have two goals: Prove that the optimization problem of computing m o (n) and m e (n) can be simplified significantly by using elementary arguments of divisibility, as well as prove that the divisibility constraints imposed by the Equation (.) allow us to compute m o (n) and m e (n) by hand. Provide more data about the behavior of m o (n) and m e (n). 2. Computation of m o (n) and m e (n) In this section we study the divisibility properties of the solutions of Equation (.). Using these properties, we were able to compute m o (n) and m e (n) for 3 n 4 and 3 n 29, respectively. Theorem 2.. Let p be an odd prime number. Let M be a positive integer and let x, x 2,..., x n be a sequence of distinct odd positive integers such that n = and < M, for all i. x i= i () If p t divides exactly k terms x,..., x k, and p t+, for all i, let a i = /p t for i =,..., k, and let A = a a 2 a k. Then p divides k i= A/a i

4 66 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA (2) If p t x j, for some j, and p t+, for all i, then p t divides another term of the sequence { }. (3) If p s x j, for some j, and p s+ > M, then p s divides at least three terms of the sequence { }. Proof. () For j > k we have that p t /x j = c j /d j where p c j and p d j. From Equation (.) we have p t = n i= p t = k i= a i + Let D = d k+ d k+2 d n, so p D. From ( k ) ADp t A = D + A a i= i n j=k+ n j=k+ c j d j. ( ) D c j d j it follows that p divides k i= A/a i (2) This follows from part. (3) If x = a p s, x 2 = a 2 p s and p s x j for j > 2, then a + a 2 0 mod p by part. Since a + a 2 is even, then a + a 2 2p so a > p (or a 2 > p) and x > p s+ > M. The following corollaries are for specific values of M: Corollary 2.2. Let p and x, x 2,..., x n be as in Theorem 2. and let M = 240. Then: () If p > 37, p for all i; (2) If p 7, p 2, for all i; (3) 8 and 29, for all i. Proof. () Assume p x. Since p 2 > M and 7p > M, we have that p divides exactly three terms, say x = p, x 2 = 3p and x 3 = 5p, so a =, a 2 = 3, and a 3 = 5. By Theorem 2., A/a i = 23 0 mod p; since p > 37, this is a contradiction. (2) Assume p 2 divides some. Then p 2 < M so p = 7,, 3. Since p 3 > M and 7p 2 > M, p 2 divides exactly three terms and, as before, p divides 23 but p < 23, which is a contradiction. (3) The case 8 follows from Theorem 2.. For p = 29 we have that p divides three or four terms. In the first case, x = a p, x 2 = a 2 p, x 3 = a 3 p, and a, a 2, a 3 {, 3, 5, 7}, since 9p > M. If {a, a 2, a 3 } = {, 3, 5},{, 3, 7}, {, 5, 7}, or {3, 5, 7}, we respectively have that A/a i = 23, 3, 47, or 7, so p does not divide A/a i. If p divides 4 terms, then a =, a 2 = 3, a 3 = 5 and a 4 = 7, and in this case A/a i = 76 0 mod p. Corollary 2.3. Let p and x, x 2,..., x n be as in Theorem 2.. Let M = 200, then 3, and 9 for all i.

5 EGYPTIAN FRACTIONS 67 We used an exhaustive computer search over the integers that comply with Theorem 2. and Corollaries 2.2, 2.3 to find all the sequences x < x 2 < < x n = m o (n) satisfying Equation (.) with odd positive integers. We also did an exhaustive computer search over the distinct even integers that comply with similar results to the ones in this section (we omit the proof because it follows along the same line of the proofs of the results of this section) to find all the sequences x < x 2 < < x n = m e (n) satisfying Equation (.) with even positive integers. In the computation of m o (n) and m e (n), the results of this section limited the amount of possible values of the s. Tables and 2 summarize our results. The column N e (n) in Table shows the number of sequences where the value m e (n) was obtained. In Table 2 the column N o (n) shows the number of sequences where the value m o (n) was obtained. The restrictions imposed on the valid values for, i n, by using Theorem 2. and Corollaries 2.2, 2.3 significantly reduced the solution search space. The search space reduction translated to speedups of up to 70 for the computation of m e (n) where 4 n 8, as seen in Figure. This allowed us to obtain results for n 29 in m o (n) and n 4 in m e (n), respectively. Table. m e (n) for 3 n 29 n m e (n) N e (n) n m e (n) N e (n) Conjectures. m e (n) is a nondecreasing function for n 3. m o (n) is a nondecreasing function for n. 3. Calculations of m o (27) and m o (29) In this section we show how to use Theorem 2. and Corollaries 2.2 and 2.3 to obtain m o (27) and m o (29). Other cases can be obtained similarly. We

6 68 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA Table 2. m o (n) for n 4 n m o (n) N o (n) n m o (n) N o (n) Figure. Execution time for the computation of m e (n) with and without the restrictions imposed by Theorem 2. and Corollaries 2.2, 2.3. The results were obtained on a workstation with one Intel(R) Xeon(R) CPU 2.3GHz, with 4MB of cache and 32GB of RAM. are going to prove that m o (27) = min max i 27 {} = 87

7 EGYPTIAN FRACTIONS 69 and m o (29) = min max {} = 87. i 29 Using the results of Section, if m o (n) < 200, then the s are in the following set: {3, 5, 7, 9,, 3, 5, 7, 2, 23, 27, 33, 35, 39, 45, 5, 55, 63, 65, 69, 75, 77, 85, 9, 99, 05, 5, 7, 9, 35, 43, 53, 6, 65, 75, 87, 89, 95}. The following is a solution of Equation (.) when n = 27: (3.) [5, 7, 9,, 3, 5, 23, 27, 39, 45, 5, 63, 65, 69, 75, 77, 85, 9, 99, 05, 5, 9, 35, 43, 53, 75, 87]. Therefore m o (27) 87. Substituting and 05 in (3.) by 2, 35, 55, 65, we obtain a solution for n = 29: + 05 = Hence, m o (29) 87. Suppose m o (n) < 87, where n {27, 29}. This implies that m o (n) 75. Using Theorem 2., if 7 appears in a minimal solution then one of the following has to appear: () 7, 53 = 3 2 7, 87 = 7, or (2) 5 = 3 7, 85 = 5 7, 9 = 9 7, 87 = 7. Therefore, since m 0 (n) 75, 7 cannot appear in a minimal solution. Hence there are 29 possible values: 3, 5, 7, 9,, 3, 5, 2, 23, 27, 33, 35, 39, 45, 55, 63, 65, 69, 75, 77, 9, 99, 05, 5, 7, 35, 43, 65, 75. This implies that m o (29) = 87 since If 3 appears in a minimal solution, then one of the following will appear: () 3, 39 = 3 3, 7 = 9 3 (2) 3, 39 = 3 3, 9 = 7 3, 7 = 9 3, 43 = 3 (3) 65 = 5 3, 9 = 7 3, 7 = 9 3 (4) 39 = 3 3, 65 = 5 3, 7 = 9 3, 43 = 3 Among the 29 possible values we count 3, 3 3,..., 3 (six appearances of 3), but the number of times that 3 can appear without getting a contradiction are 4 or 5. If 3 appears four times, then we have 29 2 = 27 possible values. Hence, the only possible solution including 39, 65, 7, and 43 is: [3, 5, 7, 9,, 5, 2, 23, 27, 33, 35, 39, 45, 55, 63, 65, 69, 75, 77, 99, 05, 5, 7, 35, 43, 65, 75], but its sum is not equal to. If 3 appears 5 times, then we have 29 = 28 possible values, as we do not have to consider 65. Hence, in the list of 28 values we choose 22 from 23 possible values (distinct from 3, 39, 9, 7, 43) and sum them. Those sums are not equal to, and therefore m o (27) = 87. The following is an example of the solutions we found for m o (27): [ 3, 5, 3, 2, 23, 27, 35, 39, 5, 55, 63, 65, 69, 75, 77, 85, 9, 99, 05, 5, 9, 35, 43, 53, 65, 75, 87].

8 70 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA 4. Solutions of n i= / = with Restrictions In [5], the authors considered egyptian fractions with restrictions. They considered solutions of Equation (.) over S(p,..., p r ) = {p α pαr r α i N 0, i =,..., r}, where N 0 is the set of nonnegative integers. In this section we consider the solutions of Equation (.) over the set S(p,..., p r ), where p,..., p r are odd primes. In some sense the simplest case is S(p, p 2, p 3 ), since Equation (.) does not have solution over S(p, p 2 ) with p and p 2 odd primes (see Theorem 2.3 in [5]). Let N n (p,..., p r ) be the set of solutions (x,..., x n ) of Equation (.) with S(p,..., p r ), where any solution contains at least one x j that is divisible by p i for i =,..., r. Note that in general N n (p,..., p r ) is not equal to the value T n (p,..., p r ) of [5]. Using Theorem 2.3 in [5], we have that N n (p, p 2, p 3 ) for (p, p 2, p 3 ) {(3, 5, 7), (3, 5, ), (3, 5, 3)}, n sufficiently large, and N n (p, p 2, p 3 ) = 0 otherwise. The algorithm to compute the solutions with restrictions employs the same backtracking strategy that was used to find the solutions for m e and m o. The main difference is that we begin the algorithm with L, a precomputed list of integers x α α r in increasing order, where x α α r = p α pαr r and α i N 0. During backtracking, our algorithm can only choose among integers in L, which significantly speeds the computation. The largest integer p α pαr r used from L during the computation of N 5 (3, 5, 7) was Theorem 4.. Let N n (p,..., p r ) be the number of solutions of Equation (.) over S(p, p 2, p 3 ). Then: () N 3 (3, 5, 7) = 2034; (2) N 5 (3, 5, 7) = ; (3) N 7 (3, 5, ) = and N n (3, 5, ) = 0 for n 5; (4) N 23 (3, 5, 3) = 63 and N n (3, 5, 3) = 0 for n 2. Remark: N 3 (3, 5, 7) can be obtained using an argument similar to the one in [3]. We decided not to proceed in that way since the result can be obtained quickly via computer. For the N 5 (3, 5, 7) case it would be very difficult to apply this method since there are too many subcases. We now illustrate the process that is going to be used in the proof of Theorem 4.3. In [] we introduced the following identity (4.) abc = ab(a + b + c) + ac(a + b + c) + bc(a + b + c) to construct new solutions of Equation (.). Using (4.2) 05 = = ,

9 EGYPTIAN FRACTIONS 7 the following solution (given in [2]) for Equation (.) in -variables over S(3, 5, 7), = , can be lifted to a new solution in 3-variables: = This process can be repeated to obtain a solution for any odd n greater that 9. In [5], the authors proved N n (3, 5, 7) c 62 k/2 for a computable c > 0 and any odd number n = 2k +. We now prove N n (p,..., p r ) for some prime numbers p,..., p r and n n 0 odd, where n 0 is an odd positive integer. Theorem 4.3. () Let p 37 be prime. Then N n (3, 5, 7, p) if and only if n is odd. (2) N n (3, 5, ) if and only if n 7 is odd. (3) N n (3, 5, 3) if and only if n 23 is odd. Proof. For the first part of the theorem we give a complete proof for the case N 3 (3, 5, 7, ). For the other cases, we give a solution and the number that we need to substitute in order to get a new solution. To prove that N 3 (3, 5, 7, ), we use the solution [3, 5, 7,, 5, 2, 27, 33, 35, 45, 2079] and 2029 = We then have that 23 = 2 = Hence [3, 5, 7,, 5, 2, 27, 33, 35, 45, 44, 693, 33 23] is a new solution of Equation (.). We can repeat this process for any odd n greater than 3. To prove that N 3 (3, 5, 7, 3), we use the following solution: [3, 5, 7, 9, 5, 2, 35, 39, 45, 49, 637] and 637 = To prove that N 3 (3, 5, 7, 7), we use the following solution: [3, 5, 7, 9, 5, 7, 2, 35, 53, 357, 595] and 595 = 5 9. To prove that N 3 (3, 5, 7, 9), we use the following solution: [3, 5, 7, 9, 5, 9, 2, 35, 95, 285, 35] and 35 = To prove that N 3 (3, 5, 7, 23), we use the following solution: [3, 5, 7, 9, 5, 2, 23, 35, 69, 5, 35] and 35 = To prove that N 3 (3, 5, 7, 29), we use the following solution: [3, 5, 7, 9, 5, 2, 25, 29, 45, 725, 3045] and 3045 = To prove that N 3 (3, 5, 7, 3), we use the following solution: [3, 5, 7, 9, 5, 2, 27, 3, 35, 953, 29295] and = To prove that N 3 (3, 5, 7, 37), we use the following solution: [3, 5, 7, 9, 5, 2, 25, 37, 45,, 6475] and 6475 = 5(5 7 37).

10 72 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA For the second part of the theorem, we find the following solution for Equation (.) in 7-variables over S(3, 5, ): = Using 65 = 3 5 = 5 = we then obtain the following solution for n = 9: = (3 5 ) = We can continue this process to obtain N n (3, 5, ) for all odd n 7. Using direct computation, we do not find any solution when n 5. For the third part of the theorem, we find the following solution for Equation (.) in 23-variables over S(3, 5, 3): [3, 5, 9, 3, 5, 25, 27, 39, 45, 65, 75, 8, 7, 25, 35, 95, 225, 243, 325, 35, 675, 25, 5795]. Using 95 = = 5 39 = and = 5795, we obtain a solution for equation (.) in 25-variables over S(3, 5, 3). We can repeat the process to obtain a solution for n 25. Using the computer we do not find any solution for n 2. In [5], the authors proved that the set of natural numbers n such that the equation (.) has at least one solution over S(p,..., p r ) is a union of finitely many arithmetic progressions. Our calculations suggest the following: Let n 0 be the smallest natural number such that N n0 (p,..., p r ). Then N n (p,..., p r ) for n = 2k + n On the calculation of lim inf x n x n k In this section we prove the existence of solutions of Equation (.) satisfying certain properties. Applying the obtained results, we prove that lim inf x n /x n k = for fixed k.

11 EGYPTIAN FRACTIONS 73 Lemma 5.. For each odd positive integer q there exists an increasing sequence x, x 2,..., x n of odd positive integers such that q x n and = x + x x n. Proof. Let y, y 2,..., y m be an increasing sequence of odd positive integers such that = m i= /y i. By using one of the identities x = (3x + )/2 + 3x + if x 3 mod 4 3x(3x + )/2 or x = (3x + 3)/2 + 3x + if x mod 4 x(3x + 3)/2 we can replace /y m by one of these sums of three distinct odd positive integers and obtain a new increasing sequence with m + 2 terms of odd positive integers such that = m+2 i= but now with a different y m. Thus, we may assume that the sequence y, y 2,..., y m also has the property that q < y m. In [2, 5], it was proved that any rational number with odd denominator is a sum of a finite number of distinct terms from the sequence /3, /5, /7,, so the fraction q/y m has an expansion of the form, q s =, y m z j where the z j are distinct odd positive integers. From here it follows that s = y m (qz j ). j= j= Notice that each term qz j is greater than y m, so after substituting this expression for /y m into the equation = m i= /y i, we obtain a sequence that satisfies the statement of the lemma. Lemma 5.2. Let k be a positive even integer and n < n 2 < < n k be a sequence of odd positive integers such that n > k + 2. Let d = n n 2 n k and q = d k i= d/n i. Then q is an odd integer, d < qn, and (5.) q = d qn qn 2 qn k Proof. Since k is an even integer and the n i are odd, the integer q is odd. Moreover, ( ) k q = k = qn i q q q d = d. i= y i, n i= i

12 74 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA so To show that d < qn notice that k n d < kd, n j This proves the lemma. qn = n d j= k j= n d n j > (n k)d > d. Theorem 5.3. Let k, n < n 2 < < n k, and q be as in Lemma 5.2. Then there exists a sequence x < x 2 < < x n of odd positive integers such that = x + x x n and whose last k terms are x n = aqn k, x n = aqn k,, x n (k ) = aqn. Proof. By Lemma 5.there exists a sequence z < z 2 < < z m such that q z m and = z z 2 z m Therefore z m = aq for some odd integer a. We now multiply both sides of (5.) by /a and then substitute /z m = /qa by this new value to give the theorem. The following is a consequence of Theorem 5.3: Corollary 5.4. Let k be a natural number. Then x n lim inf =. x n k Proof. We are going to prove the case when k = 2. By Theorem 5.3, there exists a solution of Equation (.) for each pair (n, n 2 ) = (2n +, 2n + 3), where n is a natural number greater than. This implies the corollary. 6. On max satisfying n i= / = over the even numbers In this section we compute max for sets of satisfying Equation (.) over distinct even numbers. Let E be the set of positive even numbers. In [6], Sylvester introduced the sequence a = 2, a 2 = 3, a 3 = 7,..., a n+ = a a 2 a n +, which satisfies the equation S = =. a a 2 a n a a 2 a n From [8, 6, 4], it is known that { (6.) max N n+ i= } = = a a n,

13 EGYPTIAN FRACTIONS 75 and if x x 2 x n are natural numbers that satisfy + + <, x x n then (6.2) x x n a a n In fact, Equation (6.2) implies Equation (6.). Theorem 6.. Let a,, a n be the sequence defined above. Then { } n+ max x n+ =, x < < x n+, E = 2a a n. i= Proof. Observe that S e = =. 2a 2a n 2a a n Let c = 2 and c i = 2a i for 2 i n. Then =. c c 2 c n 2a a n We claim that { (6.3) max E n+ i= } = = 2a a n. Let x, x 2,, x n+ E be a sequence such that + + = = x x n , 2a 2a a n where x < x 2 < < x n+. Then + + x x n+ 2 = + + 2a 2a a n so (x /2) + + (x n+ /2) = + + a This implies that (x /2) + + (x n+ /2) = 2. a a n =. If x = 2, then (x 2 /2) + + (x n+ /2) =. Therefore x n+ /2 a a n and hence x n+ 2a a n. Suppose x 4 and let x i = /2. Suppose there exists a subset A of B = {x,..., x n+ } such that =. x i A i

14 76 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA Then x i A i and x =, i A c i where A c = B A. Note that x i a a m, where m = max{ A, A c }. Hence 2a a m,. In particular, x n+ 2a a n, so we can assume that i A /x i for any subset A of B = {x,..., x n+ }. Suppose we take out some collection of x i from B = {x,..., x n+ } such that the sum of the reciprocal of the elements left in B is less than, but when we add x n+ to the set, we obtain a sum greater than. Call this set A and observe that x <. x i A i = Let A = k and use Equation (6.2) to obtain x + +. x i A i a a k We then have that x + x i A i K x i A x i for some natural number K. Then x > n+ = = a + + a k + K x i A x i and hence Kx n+ < x i A x i. Furthermore, we have that K x i A x i a... a n a a k, and hence x i A x i Ka a n. This implies that x n+ < a a 2 a n. Suppose now for any subset A of B satisfying x <, x i A i we have that x + x i A i x n+ <. We choose A 0 with maximal sum, i.e, for any A B satisfying x <, x i A i

15 we have EGYPTIAN FRACTIONS 77 x i A x i. x x i A i 0 We can substitute values x i of A 0 by y i such that < and y y i A i 0 y y i A i 0 + x n+. Now we can apply the method of the previous case to obtain the inequality x n+ a a n. Acknowledgment We are grateful to the referee for his/her comments. References. R. Arce-Nazario, F. Castro, and R. Figueroa, On the number of solutions of i= = in distinct odd natural numbers, J. Number Theory 33 (203), no. 6, MR R. Breush, Solution to problem e452, American Math. Monthly 6 (954), N. Burshtein, Improving solutions of k i= = with restrictions as required by Barbeau respectively by Johnson, Discrete Math. 306 (2006), no. 3, MR , The equation 9 i= = in distinct odd integers has only the five known solutions, J. Number Theory 27 (2007), no., MR Y-G. Chen, C. Elsholtz, and L-L. Jiang, Egyptian fractions with restrictions, Acta Arith. 54 (202), no. 2, MR D. R. Curtiss, On Kellogg s Diophantine Problem, Amer. Math. Monthly 29 (922), no. 0, MR R. K. Guy, Unsolved problems in number theory, third ed., Problem Books in Mathematics, Springer-Verlag, New York, MR O. Kellogg, On a diophantine equation, Amer. Math. Monthly 28 (92), G. Martin, Dense Egyptian fractions, Trans. Amer. Math. Soc. 35 (999), no. 9, MR , Denser Egyptian fractions, Acta Arith. 95 (2000), no. 3, MR P. Shiu, Egyptian fraction representations of with odd denominators, Math. Gaz. 93 (2009), W. Sierpiński, Sur les décompositions de nombres rationnels en fractions primaires, Mathesis 65 (956), MR N. Sloane and S. Plouffe, Integer Sequences A K. Soundararajan, Approximating from below using n egyptian fractions, 2005, pp B. M. Stewart, Sums of distinct divisors, Amer. J. Math. 76 (954), MR J. J. Sylvester, On a Point in the Theory of Vulgar Fractions, Amer. J. Math. 3 (880), no. 4, MR

16 78 R.ARCE-NAZARIO, F. CASTRO, AND R. FIGUEROA Department of Computer Science, University of Puerto Rico, Río Piedras Campus address: Department of Mathematics, University of Puerto Rico, Río Piedras Campus address: Department of Mathematics, University of Puerto Rico, Río Piedras Campus address: