#A26 INTEGERS 10 (2010), ON CONGRUENCE CONDITIONS FOR PRIMALITY

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1 #A6 INTEGERS 10 (010), ON CONGRUENCE CONDITIONS FOR PRIMALITY Sherry Gong Departments of Mathematics and Physics, Harvard University Scott Duke Kominers 1 Department of Economics, Harvard University, and Harvard Business School kominers@fas.harvard.edu, skominers@gmail.com Received: 5//09, Revised: /8/10, Accepted: 3/7/10, Published: 7/16/10 Abstract For any k 0, all primes n satisfy the congruence nσ k (n) mod ϕ(n). We show that this congruence in fact characterizes the primes, in the sense that it is satisfied by only finitely many composite n. This characterization generalizes the results of Lescot and Subbarao for the cases k = 0 and k = 1. For 0 k 14, we enumerate the composite n satisfying the congruence. We also prove that any composite n which satisfies the congruence for some k satisfies it for infinitely many k. 1. Introduction Lescot [1] and Subbarao [] showed that, for k {0, 1}, the congruence nσ k (n) mod ϕ(n), (1) in some sense characterizes the set P of primes. Specifically, they respectively showed the k = 0 and k = 1 cases of the following theorem. Theorem 1. For k {0, 1} and n N, the congruence (1) holds if and only if n P {1} S k, where S 0 = {4, 6, 14} and S 1 = {4, 6, }. Here, σ k and ϕ respectively denote the divisor and totient functions, defined by σ k (n) = d n d k = p (αi+1)k p k, ϕ(n) = {d N : 1 d < n and (d, n) = 1} = for k 0 and N n = p α1 1 pαr r (with the p i P distinct). 1 Corresponding author. p αi 1 i (p ),

2 INTEGERS: 10 (010) 314 It is clear that if n P, then σ k (n) = n k + 1 and ϕ(n) = n 1. In this case, we have nσ k (n) n(n k + 1) mod (n 1), as p 1 mod (p 1). We therefore see that, for any k 0, all n P satisfy (1). In this note, we present the following result generalizing the reverse direction of Theorem 1. Theorem. For any k 0, let S k be the set of composite n N satisfying (1). Then, (i) S k P, (ii) S k is finite, and (iii) the maximal element of S k is at most k We prove Theorem in Section. Then, in Section 3, we enumerate the sets S k for 0 k 14. There, we also prove that any n N which appears in S k for some k 0 appears in infinitely many of the sets {S k } k =0.. Proof of Theorem We suppose that n S k, i.e., that n is composite and satisfies (1). Upon writing n = α p α1 1 pαr r with the p i P distinct and odd, we have ϕ(n) = α 1 p αi 1 i (p ). () For any i (1 i r) such that α i > 1, we see from the expression () that p i ϕ(n). Since clearly p i n, the congruence (1) then gives that p i an impossibility. Thus, we must have α i = 1. An analogous argument shows that α {1, }. Now, as p is even for each i (1 i r), we see from () that r ϕ(n). Furthermore, the expression σ k (n) = (α+1)k 1 = (α+1)k 1 p (αi+1)k p k p k p k = (α+1)k 1 (p k i + 1) (3)

3 INTEGERS: 10 (010) 315 yields that r σ k (n) since p k i + 1. We then obtain from the congruence (1) that r, whence we see that r 1. Since n is composite, we are left with only two possibilities: n = p (with p P) or n = 4p (with p P). The second case is impossible, as when n = 4p we have (1) and 4 ϕ(n), together implying 4 ; the fact that S k P follows. In the first case, if p =, then n = 4 k If p is odd, then we have nσ k (n) = p( k +1)(p k +1) and ϕ(n) = p 1. Because p 1 mod (p 1), we see that nσ k (n) p( k + 1)(p k + 1) 4( k + 1) mod (p 1). (4) Combining (1) and (4), we see that (p 1) k+ +. (5) As n = p and (5) implies that p k+ + 3, we have the stated bound on the size of n S k ; the finitude of S k follows. 3. The Sets S k Table 1 presents the exceptional sets S k for 0 k 14. It is clear that 4, 6 S k for each k 0, since ϕ(4) = ϕ(6) =, and 4σ k (4) and 6σ k (6) are even for all k. Beyond this observation, however, the behavior of the sets S k appears to be quite erratic. Nonetheless, we obtain the following partial characterization result for the S k. Corollary 3. If n N is in S k for some k 0, then it is in infinitely many of the sets {S k } k =0. Proof. It is easily seen in the proof of Theorem that n S k if and only if n = p for p P satisfying (p 1) ( k+ + ), or equivalently, p 1 ( k+1 + 1). But this means that k+1 1 mod p 1, hence k+ 1 mod p 1. It follows that we have (j+1)(k+1) 1 = (k+)j+k k mod p 1

4 INTEGERS: 10 (010) 316 k S k 0 {4,6,14} 1 {4,6,} {4,6,14,38} 3 {4,6} 4 {4,6,14,46,134} 5 {4,6,,6} 6 {4,6,14} 7 {4,6} 8 {4,6,14,38} 9 {4,6,,166} 10 {4,6,14,734} 11 {4,6} 1 {4,6,14} 13 {4,6,,118,454} 14 {4,6,14,38,46,134,398,3974,14566} Table 1: The exceptional sets S k (0 k 14) for any j N. Then, for any k {(j + 1)(k + 1) 1} j=1, we have (p 1) ( k + + ), hence n S k. We have therefore produced infinitely many k such that n S k. References [1] P. Lescot, A characterisation of prime numbers, The Mathematical Gazette 80 (1996), no. 488,

5 INTEGERS: 10 (010) 317 [] M. V. Subbarao, On two congruences for primality, Pacific Journal of Mathematics 5 (1974),