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Cullen Numbers with the Lehmer Property 131 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 140, Number 1, January 2012, Pages 129–134 S 0002-9939(2011)10899-2 Article electronically published on May 20, 2011 CULLEN NUMBERS WITH THE LEHMER PROPERTY JOSEMAR´ ´IA GRAU RIBAS AND FLORIAN LUCA (Communicated by Matthew A. Papanikolas) Abstract. Here, we show that there is no positive integer n such that the n nth Cullen number Cn = n2 + 1 has the property that it is composite but φ(Cn) | Cn − 1. 1. Introduction n A Cullen number is a number of the form Cn = n2 +1 forsome n ≥ 1. They attracted the attention of researchers since it seems that it is hard to find primes of this form. Indeed, Hooley [8] showed that for most n the number Cn is composite. For more about testing Cn for primality, see [3] and [6]. For an integer a>1, a pseudoprime to base a is a composite positive integer m such that am ≡ a (mod m). Pseudoprime Cullen numbers have also been studied. For example, in [12], it is shown that for most n, Cn is not a base a pseudoprime. Some computer searches up to several millions did not turn up any pseudoprime Cn to any base. Thus, it would seem that Cullen numbers which are pseudoprimes are very scarce. A Carmichael number is a positive integer m which is a base a pseudoprime for any a. A composite integer m is called a Lehmer number if φ(m) | m − 1, where φ(m) is the Euler function of m. Lehmer numbers are Carmichael numbers, hence, pseudoprimes in every base. No Lehmer number is known, although it is known that there are no Lehmer numbers in certain sequences, such as the Fibonacci sequence (see [9]), or the sequence of repunits in base g for any g ∈ [2, 1000] (see [4]). For other results on Lehmer numbers, see [1], [2], [11], [13], [14]. Our result here is that there is no Cullen number with the Lehmer property. Hence, if φ(Cn) | Cn − 1, then Cn is prime. Theorem 1. Let Cn be the nth Cullen number. If φ(Cn) | Cn − 1,thenCn is prime. 2. Proof of Theorem 1 Assume that n ≥ 30, that φ(Cn) | Cn − 1, but that Cn is not prime. Then Cn is square-free. Write k Cn = pi. i=1 Received by the editors October 14, 2010 and, in revised form, November 11, 2010. 2010 Mathematics Subject Classification. Primary 11A05; Secondary 11N25, 11A07. c 2011 American Mathematical Society 129 Licensed to University de Oviedo. Prepared on Thu Nov 15 06:32:23 EST 2012 for download from IP 156.35.62.18/156.35.192.4. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 130 JOSEMAR´ ´IA GRAU RIBAS AND FLORIAN LUCA So, k n (pi − 1) | n2 . i=1 α n2 Write n =2 n1,wheren1 is odd. Then Cn = n12 +1,wheren2 := α + n.Letp np be any prime factor of Cn.Sincep − 1 | Cn − 1, it follows that p = mp2 +1for some odd divisor mp of n and some np with log n n ≤ n = n + α ≤ n + . p 2 log 2 Letusfirstshowthatinfactnp ≤ n. Assume that np >n. Then, n (1) Cn = n2 +1=pλ, n+1 for some positive integer λ,wherep ≥ 2 +1.Observethatλ>1 because Cn is not prime. Now C n2n +1 λ = n ≤ <n. p 2n+1 +1 Reducing equation (1) modulo 2n,wegetthat2n | λ − 1, so 2n ≤ λ − 1 <n,which is false for any n>1. Hence, np ≤ n. np Next we look at mp.Ifmp =1,thenp =2 + 1 is a Fermat prime. Hence, γp γp np =2 for some nonnegative integer γ. Since 2 = np ≤ n,wegetthatγp < (log n)/(log 2). Hence, the prime p can take at most 1+(log n)/(log 2) values. Next, observe that since (2) mp | n, p|Cn it follows that the number of prime factors p of Cn such that mp > 1is≤ (log n)/(log 3). Hence, we have arrived at the bound log n log n (3) k<1+ + < 1+2.4logn. log 2 log 3 We next bound np.PutN := n/ log n, and consider pairs (a, b) of integers in {0, 1,...,N}.Thereare(N +1)2 >n/log n such pairs. For each such pair, 3/2 1/2 consider the expression L(a, b):=an + bnp ∈ [0, 2n /(log n) ]. Thus, there exist two pairs (a, b) =( a1,b1) such that 2n3/2/(log n)1/2 |(a − a )n +(b − b )n | = |L(a, b) − L(a ,b )|≤ < 3(n log n)1/2. 1 1 p 1 1 n/ log n − 1 Put u := a − a1,v:= b − b1.Then(u, v) =(0 , 0) and 1/2 |un + vnp| < 3(n log n) . We may also assume that u and v are coprime, for if not, we replace the pair (u, v) by the pair (u1,v1), where d := gcd(u, v),u1 := u/d, v1 := v/d,andthe 1/2 1/2 properties that max{|u1|, |v1|} ≤ (n/ log n) and |u1n + v1np| < 3(n log n) are still fulfilled. Finally, up to replacing the pair (u, v) by the pair (−u, −v), we may assume that u ≥ 0. Licensed to University de Oviedo. Prepared on Thu Nov 15 06:32:23 EST 2012 for download from IP 156.35.62.18/156.35.192.4. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use CULLEN NUMBERS WITH THE LEHMER PROPERTY 131 n np Now consider the congruences n2 ≡−1(modp)andmp2 ≡−1(modp). Observe that 2,n,mp are all three coprime to p. Raise the first congruence to u and the second to v and multiply them to get u v nu+npv ≡ − u+v n mp2 ( 1) (mod p). Hence, p divides the numerator of the rational number u v nu+npv − − u+v (4) A := n mp2 ( 1) . α Let us show that A = 0. Assume that A = 0. Recall that n =2n1.Thus, expression (4) is u v (n+α)u+npv − − u+v A = n1 mp2 ( 1) =0. u v ≥ Then n1 mp =1,(n + α)u + vnp =0,andu + v is even. Since u 0, it follows that v ≤ 0. Put w := −v,sow ≥ 0. There exists an odd positive integer ρ such that w u n1 = ρ and mp = ρ .Sinceu and v are coprime and u + v is even, it follows that u and v are both odd. Hence, w is also odd. Also, since mp divides n1, it follows that u ≤ w.Wenowget α w (2 ρ + α)u − wnp =0, so u w = α w . np 2 ρ + α α u α u The left-hand side is ≥ u/n = u/(2 ρ ), because np ≤ n =2 ρ . Hence, we get that u ≤ u w u ≤ w ≤ w α u = α w leading to u w α w . 2 ρ np 2 ρ + α ρ ρ +(α/2 ) ρ For ρ ≥ 3, the function s → s/ρs is decreasing for s ≥ 0, so the above inequality together with the fact that u ≤ w implies that u = w (sobothare1becausethey are coprime), and that all the intermediary inequalities are also equalities. This means that u = w =1,α =0andn = np, but all this is possible only when Cn = p, which is not allowed. If ρ =1,wethengetthatn1 = 1, so every prime factor p of Cn is a Fermat prime. Hence, we get k γp γp n2 2 i ∈ 2 i Cn =2 +1= (2 +1)= 2 i I , i=1 I⊆{1,...,k} and k ≥ 2, but this is impossible by the unicity of the binary expansion of Cn. Thus, it is not possible for the expression A shown at (4) to be zero. The size of the numerator of A is at most | | | | 1/2 1/2 1+ nu+npv u v ≤ 1+3(n log n) 2(n/ log n) 2 n mp 2 n 1/2 1/2 1/2 < 21+3(n log n) +(2/ log 2)(n log n) < 26(n log n) . In the above chain of inequalities, we used the fact that 3 + 2/ log 2 < 5.9, together with the fact that (n log n)1/2 > 10 for n ≥ 30. Thus, for n ≥ 30, we have that the inequality 1/2 (5) p<26(n log n) holds for all prime factors p of Cn. Licensed to University de Oviedo. Prepared on Thu Nov 15 06:32:23 EST 2012 for download from IP 156.35.62.18/156.35.192.4. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 132 JOSEMAR´ ´IA GRAU RIBAS AND FLORIAN LUCA Thus, we get the inequality k k n 6(n log n)1/2 6k(n log n)1/2 2 <Cn = pi < 2 =2 , i=1 i=1 leading to n1/2 (6) k> . 6(log n)1/2 Comparing estimates (3) and (6), we get n1/2 < 1+2.4logn, 6(log n)1/2 implying that n<6 × 105. It remains to lower this bound. We first lower it to n<93000. Indeed, first 5 2γ note that since n<6 × 10 , it follows that if p = Fγ =2 + 1 is a Fermat prime dividing Cn,thenγ ≤ 18. The only such Fermat primes are for γ ∈{0, 1, 2, 3, 4}. Furthermore, (log n)/(log 3) ≤ log(6 × 105)/(log 3) = 12.1104 ...
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