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Comptes RendusMathématiqueJin-Hui Fang and Fang-Gang XueOn weakly prime-additive numbers with length 4k +3Volume 362 (2024), p. 275-278Online since: 2 May 2024https://doi.org/10.5802/crmath.555This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/C EN T R EMER S ENN EThe Comptes Rendus. Mathématique are a member of theMersenne Center for open scientific publishingwww.centre-mersenne.org — e-ISSN : 1778-3569Comptes Rendus. Mathématique2024, Vol. 362, p. 275-278https://doi.org/10.5802/crmath.555Research article / Article de rechercheNumber theory / Théorie des nombresOn weakly prime-additive numbers withlength 4k +3Jin-Hui Fang ∗,a and Fang-Gang Xue ba School of Mathematical Sciences, Nanjing Normal University, Nanjing 210023, P.R. Chinab Nanjing University of Information Science & Technology, Nanjing 210044, P.R. ChinaE-mails: fangjinhui1114@163.com, xfg950118@126.comAbstract. If a positive integer n has at least two distinct prime divisors and can be written as n = pα11 +·· ·+pαtt ,where p1 < ·· · < pt are prime divisors of n and α1, . . . ,αt are positive integers, then we define such n asweakly prime-additive. Obviously, t ⩾ 3. Following Erdős and Hegyvári’s work, Fang and Chen [J. NumberTheorey 182(2018), 258-270] obtained the following result: for any positive integer m, there exist infinitelymany weakly prime-additive numbers n with m | n and n = pα11 +·· ·+pα55 , where p1, . . . , p5 are distinct primedivisors of n and α1, . . . ,α5 are positive integers. In this paper, we prove the existence of such n with generallength t , where t ≡ 3 (mod 4) and t > 3. The main result is summarized as follows: for any positive integersm, t with t ≡ 3 (mod 4) and t > 3, there exist infinitely many weakly prime-additive numbers n with m | n andn = pα11 +·· ·+pαtt , where p1, . . . , pt are distinct prime divisors of n and α1, . . . ,αt are positive integers.Keywords. weakly prime-additive numbers, Dirichlet’s theorem, the Chinese remainder theorem.2020 Mathematics Subject Classification. 11A07, 11A41.Funding. This work is supported by the National Natural Science Foundation of China, Grant No. 12171246and the Natural Science Foundation of Jiangsu Province, Grant No. BK20211282.Manuscript received 28 July 2023, revised 14 August 2023, accepted 17 August 2023.1. IntroductionIf a positive integer n has at least two distinct prime divisors and can be written as n = pα11 +·· · + pαtt , where p1 < ·· · < pt are prime divisors of n and α1, . . . ,αt are positive integers, thenwe define such n as weakly prime-additive. Obviously, t ⩾ 3. In 1992, Erdős and Hegyvári [2]firstly considered weakly prime-additive numbers and proved that, for any prime p, there existinfinitely many weakly prime-additive numbers n such that p|n.In 2018, Fang and Chen [5] made further research on weakly prime-additive numbers andobtained the following result:Theorem 1. For any positive integer m, there exist infinitely many weakly prime-additive num-bers with t = 3 which are divisible by m if and only if 8 ∤m.In [5], Fang and Chen also proved that:∗Corresponding author.ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/276 Jin-Hui Fang and Fang-Gang XueTheorem 2. For any positive integer m, there exist infinitely many weakly prime-additive num-bers n with m | n andn = pα11 +·· ·+pα55 ,where p1, . . . , p5 are distinct prime divisors of n and α1, . . . ,α5 are positive integers.Afterwards, Fang [3] proved that, for any positive integers m, t with t ≡ 1 (mod 4) and t > 1,there exist infinitely many weakly prime-additive numbers n with m | n and n = pα11 + ·· ·+pαtt ,where p1, . . . , pt are distinct prime divisors of n and α1, . . . ,αt are positive integers. In this paper,by adding some tricky arguments, we solve the case t ≡ 3 (mod 4), where t > 3, that is:Theorem 3. For any positive integers m, t with t ≡ 3 (mod 4) and t > 3, there exist infinitely manyweakly prime-additive numbers n with m | n andn = pα11 +·· ·+pαtt , (1)where p1, . . . , pt are distinct prime divisors of n and α1, . . . ,αt are positive integers.Let k, l be integers with k ⩾ 2 and l ⩾ 3. WriteSk ={n : n = ∑p|npk ,ω(n)⩾ 2},where ω(n) is the number of distinct prime factors of n. In 2005, De Koninck and Luca [1]obtained many nice results about Sk . As a main result, they considered the nature of Sk andidentified all integers n in S3 withω(n) = 3. Let {Pm}m⩾0 denote the Pell sequence given by P0 = 0,P1 = 1 and Pm+1 = 2Pm +Pm−1 for m ⩾ 1. In 2022, Fang [4] proved that, a positive integer n canbe expressed as n = 22 + p2 + q2, where p, q are distinct odd prime factors of n, if and only if,n = 22 +P 22m+1 +P 22m+3 for some positive integer m, where P2m+1, P2m+3 are both primes. Onemay refer to [1] and [4] for details.2. Proof of Theorem 3The idea is from [5, Theorem 1.3] and [3, Theorem 1].Proof. Write t = 4k+3 and m = 2am1, where a ⩾ 0 and 2 ∤m1. Then k ⩾ 1. Letφ denote the Eulertotient function. Similar to [3, Theorem 1], we consider the integersn = 2u +pα11 +pα22 +·· ·+pα4k+14k+1 +p4k+2, (2)where pi are primes fixed later,u = (a +2)φ(m1)4k+2∏i=1(pi −1) and αi =4k+2∏j=i+1p j −12for i = 1,2, . . . ,4k +1.(The method of this proof and the construction are the same as [3, Theorem 1], but we use a trickyidea during the choices of p1, p2, . . . , p4k+2, which is the key point during the proof.)By the Chinese remainder theorem and Dirichlet’s theorem there exists a prime p1 >max{2a+2,m1,5} such thatp1 ≡ 1 (mod 2a+2), p1 ≡−1 (mod m1).By the Chinese remainder theorem and Dirichlet’s theorem there exists a prime p2 > p1 such thatp2 ≡−1 (mod 2a+2), p2 ≡ 1 (mod m1), p2 ≡ 1 (mod p1).By the Chinese remainder theorem and Dirichlet’s theorem there exists a prime p3 > p2 such thatp3 ≡−1 (mod 2a+2), p3 ≡ 1 (mod m1), p3 ≡−1 (mod p1), p3 ≡ 1 (mod p2).Jin-Hui Fang and Fang-Gang Xue 277By the Chinese remainder theorem and Dirichlet’s theorem there exists a prime p4 > p3 such thatp4 ≡−1 (mod 2a+2), p4 ≡−1 (mod m1), p4 ≡ 1 (mod p1), p4 ≡−1 (mod p2), p4 ≡ 1 (mod p3).For i ⩾ 5, we will choose pi successively according to the parity of i . For j ⩾ 3, we firstly choosep2 j−1, where 3 ⩽ j ⩽ 2k. By the Chinese remainder theorem and Dirichlet’s theorem there existsa prime p2 j−1 > p2 j−2 such thatp2 j−1 ≡−1 (mod 2a+2), p2 j−1 ≡−1 (mod m1),p2 j−1 ≡−1 (mod p1), p2 j−1 ≡ 1 (mod p2), p2 j−1 ≡ 1 (mod p3), p2 j−1 ≡ 1 (mod p4),p2 j−1 ≡ 1 (mod p2s−1), p2 j−1 ≡−1 (mod p2s ) for s = 3,4, . . . , j −1.(If j = 3, then we omit the last procedure).Now we choose p2 j . For 3 ⩽ j ⩽ 2k, by the Chinese remainder theorem and Dirichlet’stheorem there exists a prime p2 j > p2 j−1 such thatp2 j ≡−1 (mod 2a+2), p2 j ≡ 1 (mod m1),p2 j ≡ 1 (mod p1), p2 j ≡−1 (mod p2), p2 j ≡−1 (mod p3), p2 j ≡−1 (mod p4),p2 j ≡−1 (mod p2s−1), p2 j ≡ 1 (mod p2s ) for s = 3,4, . . . , j −1, p2 j ≡−1 (mod p2 j−1).Furthermore, we add the restrictionp2 ≡ p3 ≡ 1 (mod 3), p4 ≡ 2 (mod 3), p2 j ≡ 1 (mod 5) for j = 3,4, . . . ,2k. (3)(If 3 | m1, then the congruences p2 ≡ p3 ≡ 1 (mod 3) and p4 ≡ 2 (mod 3) follow from the conditionp2 ≡ p3 ≡ 1 (mod m1) and p4 ≡ −1 (mod m1); if 5 | m1, then the congruences p2 j ≡ 1 (mod 5)follow from the condition p2 j ≡ 1 (mod m1) for j = 3,4, . . . ,2k; if (3,m1) = 1 or (5,m1) = 1, then (3)follows from Dirichlet’s theorem.)By the Chinese remainder theorem and Dirichlet’s theorem there exists a prime p4k+1 > p4ksuch thatp4k+1 ≡−1 (mod 2a+2), p4k+1 ≡ 1 (mod m1),p4k+1 ≡−1 (mod p1), p4k+1 ≡ 1 (mod p2), p4k+1 ≡ 1 (mod p3), p4k+1 ≡ 1 (mod p4),p4k+1 ≡ 1 (mod p2s−1), p4k+1 ≡−1 (mod p2s ) for s = 3,4, . . . ,2k.Finally, we will choose p4k+2. By the Chinese remainder theorem and Dirichlet’s theorem thereexists a prime p4k+2 > p4k+1 such thatp4k+2 ≡ 4k −1 (mod 2a+2), p4k+2 ≡−2 (mod m1),p4k+2 ≡−1 (mod p1), p4k+2 ≡−3 (mod p2), p4k+2 ≡−3 (mod p3),p4k+2 ≡−3 (mod p4), p4k+2 ≡ 1 (mod p5),p4k+2 ≡−5 (mod p2s ), p4k+2 ≡ 1 (mod p2s+1) for s = 3,4, . . . ,2k.(If k = 1, then we delete the middle stage and choose p5 and p6 by the last process).Obviously,p4k+2 > p4k+1 > ·· · > p2 > p1 > max{2a+2,m1,5}.Noting that p1 ≡ 1 (mod 4), p4k+2 ≡ 4k −1 ≡ 3 (mod 4) and pi ≡ 3 (mod 4) (i = 2, . . . ,4k +1), wecould deduce from the law of quadratic reciprocity that(p1p j)=(p jp1)for j = 2, . . . ,4k +2,(pip j)=−(p jpi)for 2⩽ i < j ⩽ 4k +2. (4)It follows from the definition of αi andp12 (p j −1)i ≡(pip j)(mod p j ), 1⩽ i < j ⩽ 4k +2278 Jin-Hui Fang and Fang-Gang Xuethatpαii ≡(pip j)(mod p j ), 1⩽ i < j ⩽ 4k +2.Thus, for each positive integer n with the form (2), we could deduce from (4) and the fact αi isodd thatn ≡ 0+1+ (−1)+·· ·+ (−1)︸ ︷︷ ︸4k times+(4k −1) ≡ 0 (mod 2a+2),n ≡ 1+ (−1)+1+1+ (−1)+ (−1)+1+·· ·+ (−1)+1+1+ (−2) ≡ 0 (mod m1),n ≡ 1+0+1+ (−1)+1+ (−1)+·· ·+1+ (−1)+ (−1) ≡ 0 (mod p1),n ≡ 1+(p1p2)+0+1+ (−1)+·· ·+1+ (−1)+1+ (−3) ≡ 0 (mod p2),n ≡ 1+(p1p3)+(p2p3)+0+1+1+ (−1)+·· ·+1+ (−1)+1+ (−3) ≡ 0 (mod p3),n ≡ 1+(p1p4)+(p2p4)+(p3p4)+0+1+ (−1)+·· ·+1+ (−1)+1+ (−3) ≡ 0 (mod p4).For j ⩾ 3, we firstly consider p2 j−1, where 3⩽ j ⩽ 2k +1, we haven ≡ 1+(p1p2 j−1)+·· ·+(p2 j−2p2 j−1)+0+ (−1)+1+·· ·+ (−1)+1+1 ≡ 0 (mod p2 j−1).Now we consider p2 j , where 3⩽ j ⩽ 2k, we haven ≡ 1+(p1p2 j)+·· ·+(p2 j−1p2 j)+0+ (−1)+1+·· ·+1+ (−1)+ (−5) ≡ 0 (mod p2 j ).Finally, we consider p4k+2, we could deduce from (3) and (4) thatn ≡ 1+(p1p4k+2)+(p2p4k+2)+·· ·+(p4kp4k+2)+(p4k+1p4k+2)≡ 1+1−( p23)−( p33)−( p43)+ (−1)+( p65)+·· ·+ (−1)+( p4k5)+ (−1)≡ 0 (mod p4k+2).To sum up,n = 2u +pα11 +pα22 +·· ·+pα4k+14k+1 +p4k+2and 4mp1p2 . . . p4k+1p4k+2 = 2a+2m1p1p2 . . . p4k+1p4k+2 | n.This completes the proof of Theorem 3. □AcknowledgmentsThe authors sincerely thank the referees for their invaluable suggestions which improve thequality of this manuscript to a large extent.References[1] J.-M. De Koninck, F. Luca, “Integers representable as the sum of powers of their prime factors”, Funct. Approximatio,Comment. Math. 33 (2005), p. 57-72.[2] P. Erdős, N. Hegyvári, “On prime-additive numbers”, Stud. Sci. Math. Hung. 27 (1992), no. 1-2, p. 207-212.[3] J.-H. Fang, “Note on the weakly prime-additive numbers”, J. Nanjing Norm. Univ., Nat. Sci. Ed. 41 (2018), no. 4, p. 26-28.[4] ——— , “A note on weakly prime-additive numbers”, Int. J. Number Theory 18 (2022), no. 1, p. 175-178.[5] J.-H. Fang, Y.-G. Chen, “On the shortest weakly prime-additive numbers”, J. Number Theory 182 (2018), p. 258-270.

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