Annaes Mathematicae et Informaticae 33 (2006 pp. 45 56 http://www.ektf.hu/tanszek/matematika/ami On prime divisors of remarkabe sequences Ferdinánd Fiip a, Kámán Liptai b1, János T. Tóth c2 a Department of Mathematics University of J. Seye e-mai: fiip.ferdinand@seznam.cz b Institute of Mathematics and Informatics Eszterházy Károy Coege e-mai: iptaik@ektf.hu c Department of Mathematics University of Ostrava e-mai: toth@osu.cz Submitted 10 November 2006; Accepted 18 December 2006 Abstract In this paper we study sequences of the form (a n + b n=1, where a, b N. We prove many interesting resuts connection with sequences with infinitey many prime divisors. Keywords: prime divisors, Dirichet s theorem MSC: 11N13 1. Introduction There are many mathematica probems when we investigate the divisibiity of sequences by a prime. We usuay find this kind of interesting exampes in nationa mathematica competitions and in the Internationa Math Oympiad. In this paper we study sequences of the form (a n + b n=1, where a,b N. We prove some resuts concerning with sequences with infinitey many prime divisors. Moreover we characterize these sequences. Some of our theorems assert that there are infinitey many prime divisors of a sequence. These statements come from easiy from the theory of S-units, but in this paper we use ony eementary methods to get our resuts. We mention that our resuts hep to generaize probems which can be found in some exercise books for students. 1 Research supported by the Hungarian Nationa Foundation for Scientific Research Grant. No. T 048945 MAT 2 Research supported by Grant ČR 201/04/0381/2 45
46 F. Fiip, K. Liptai, J. T. Tóth Let A = {a 1 < a 2 < < a n < } N be a given set and et us denote by A(x the number of the eements of A not exceeding x. Let us suppose for any natura number k there is a positive rea number x k such that for a x > x k the inequaity A(x > (og x k hods. In this case there are infinitey many different prime divisors of the eements of A (see [3], p. 102. Further we sha study the sequences of positive integers where the previous condition is not true. Let a,b be natura numbers with a > 1 and (a,b = 1. Obviousy the sequences (a n + b n=1 (1.1 do not fufi the above condition, since [ ] og(x b A(x = og a if x > b + 1. In what foows we show that sequences (1.1 have infinitey many different prime divisors. In the specia case, when a = 10 and b = 3 we proved (in [6] that the sequence (10 n + 3 n=1 has infinitey many prime divisors, moreover for infinitey many primes p there are infinitey many n N such that p 10 n + 3. 2. Resuts First we prove that there are subsequences of sequences (1.1 which have infinitey many prime divisors. Theorem 2.1. Let a, b, c, d be natura numbers, (a, b = 1 and a > 1. Then there are infinitey many prime divisors of the sequences (a c+(n 1d + b n=1. (2.1 Proof. First we suppose that sequence (2.1 has ony finitey many prime divisors. Let us denote these primes by q 1 < q 2 < < q k. Let us denote by q 1 < q 2 < < q the prime divisors of sequence (2.1 which are divisors of a c +b as we and q +1 < q +2 < < q k which are not divisors of a c + b. Let us denote by α s for a 1 s and s N the east natura number such that Let q αs s > a c + b. M = q α1 1 qα2 2 qα q +1 q +2 q k be a product of prime powers. In this case (a,m = 1 since (a,b = 1. By the theorem of Euer we have M a nϕ(m 1 (2.2 for a n N.
On prime divisors of remarkabe sequences 47 Now we investigate the sequence which is obviousy a subsequence of sequence (2.1. Let q be a prime divisor of sequence (2.3 that is for some m N. It foows from (2.2 that Using (2.4 and (2.5 we have (a c+mϕ(md + b m=1 (2.3 a mϕ(md+c + b 0 (mod q (2.4 a mϕ(md 1 0 (mod q. (2.5 a mϕ(md+c + b = a mϕ(md (a c 1 + a mϕ(md 1 + b + 1 a c + b (mod q. It is cear that q a c + b, it foows that q {q 1,q 2,...,q }, that is a mϕ(md+c + b = q βm 1 1 q βm 2 2 q βm where β mj 0 for a m N and 1 j. We show that for a m N and 1 j we have β mj < α j. Let 1 j, m be arbitrary natura numbers and β mj α j then q αj j a mϕ(md+c + b, that is Since q αj j a mϕ(md+c + b 0 (mod q αj j. M, it foows from (2.2 that a mϕ(md 1 0 (mod q αj j and a mϕ(md+c + b = a mϕ(md (a c 1 + a mϕ(md 1 + b + 1 a c + b (mod q αj j, that is q αj j a c + b, which is contradiction since q αj j terms of (2.3 we have > a c + b. It foows that for a a mϕ(md+c + b < q α1 1 qα2 2 qα M. In this way we obtained a contradiction since sequence (2.3 is not bounded. In the seque we prove an interesting property of the prime divisors of sequence (2.1. Theorem 2.2. If m N is a divisor of a term of sequence (2.1 then m divides infinitey many terms of sequence (2.1.
48 F. Fiip, K. Liptai, J. T. Tóth Proof. Let m N be a divisor of a term of sequence (2.1. Let us denote by n 0 the east non-negative number which m a c+n0d + b. (2.6 Since (a,m = 1, there exists a power h m of a (mod m. The number m divides a n 1 if and ony if h m n. Let us consider the sequence where (a n kd+c + b n=1 (2.7 h m n k = (k 1 (h m,d + n 0. Obviousy sequence (2.7 is a subsequence of sequence (2.1. We show that m divides ony those terms of sequence (2.1 which are the terms of (2.7 as we. a First we prove that m divides a terms of sequence (2.7. Obviousy we have a n kd+c + b = a n kd+c + b a n0d+c + a n0d+c = = a n0d+c (a (n k n 0d 1 + a n k0+c + b = = a n0d+c( hmd (k 1 a (hm,d 1 + a n 0d+c + b. (2.8 Using that d (h m,d is an integer number and the definition of h m we have a (k 1 d (hm,d hm 1 0 It foows that a nkd+c + b a n0d+c + b that is m divides a terms of (2.7. (mod m. (mod m, b Secondy we prove that if m divides a term of sequence (2.1 then this term is a term of sequence (2.7. Let us choose n N such that m a nd+c1 +b. Obviousy n n 0. Then we have m a nd+c1 + b ( a n0d+c1 + b = a n0d+c1( a d(n n0 1. Since (a,m = 1, therefore m a d(n n0 1. Using the definition of h m we have h m d(n n 0, and n = (k 1 h m d + n 0 (2.9 for some k N. From equation (2.9 we deduce ( n = (k 1 h m (h m,d d (h m,d + n 0.
On prime divisors of remarkabe sequences 49 ( Using that fraction k 1 d (hm,d h m (h m,d, d (h m,d = 1, we have that n is an integer if and ony if the is aso an integer. Consequenty d k 1 = ( 1 (h m,d, and Now the theorem is proved. h m n = ( 1 (h m,d + n 0. In the previous theorems we investigated such subsequences of sequences (1.1 where the powers formed arithmetic progressions. It is known that the asymptotic density of sets of terms of arithmetic progressions are greater than zero, more exacty it equas the reciproca of the difference. This means that sequence (2.1 is such a subsequence of (1.1 which contains reativey many terms of sequence (1.1. In what foows we are ooking for subsequences of (1.1 where the density of the set of powers is zero, but they have infinitey many prime divisors. We give two sequences possessing the above conditions. In one of them the powers run through the set of primes and in the other the powers equa the vaues of Euer s function ϕ. It is known fact that the asymptotic density of the set of primes and the set of vaues of Euer s function are zero. Theorem 2.3. Let a,b be natura numbers with (a,b = 1 and a > 1. Let us denote by p n the n-th prime number. Then the sequence has infinitey many prime divisors. (a pn + b n=1 (2.10 Proof. Let us suppose that sequence (2.10 has ony finitey many prime divisors, namey q 1,q 2,...,q k. We discuss two cases. We consider first that there are prime divisors of the terms of sequence (2.10 which divide a + b. Let us denote by q 1 < < q the divisors of a + b and q +1 < < q k which are not divisors of a + b. Let us denote by α s for a 1 s the east natura number which Put q αs s > a + b. M = q α1 1 qα2 2 qα q +1 q k. In this case (a,m = 1 since (a,b = 1. It foows from Euer s theorem that a nϕ(m 1 0 (mod M (2.11
50 F. Fiip, K. Liptai, J. T. Tóth for a n N. Using the theorem of Dirichet we get that there are infinitey many prime numbers in the sequence (nϕ(m+1 n=1. Let us denote these prime numbers by p 1 < p 2 < p n <. Obviousy the sequence (a p n + b n=1 (2.12 is a subsequence of sequence of (2.10. Let q be a prime divisor of sequence of (2.12. Obviousy q {q 1,q 2,...,q k }, moreover for some i N. It foows from (2.11 and (2.13 that a p i + b 0 (mod q (2.13 0 a p i + b a p i 1 (a 1 + a p i 1 + b a + b (mod q. Thus q a + b and q {q 1,q 2,...,q }. In other words a p i + b can be written in the form a p β i + b = q i,1 1 q βi,2 2 q β i, where β i,j 0 for a 1 j natura numbers. Now we show that β i,j < α j for a 1 j. If β i,j α j for some 1 j then a p α i + b 0 (mod q j j moreover using (2.11 and q αj j M we have a p i 1 1 It foows from the previous congruence that (mod q αj j. 0 a p i + b a p i 1 (a 1 + a p i 1 + b a + b (mod q αj j which contradicts the fact that q αj j > a + b. In this way we get a p i + b < q α 1 1 qα2 2 qα M for a i N. Here we have obtained a contradiction since sequence (2.12 is not bounded. In the second case we study when the terms of sequence (2.10 do not have such prime divisors which divide a + b. Put Since (a, L = 1, therefore L = q 1 q 2 q k. a nϕ(l 1 0 (mod L (2.14 for a n N. Let Q = ϕ(l + 1
On prime divisors of remarkabe sequences 51 be a prime and q be a prime divisor of a Q + b. It foows from the definition of Q and from (2.14 that where q {q 1,q 2,...,q k }. Obviousy a Q 1 1 (mod q 0 a Q + b a Q 1 (a 1 + a Q 1 + b a + b (mod q, which contradicts the fact that q is not a divisor of a + b. It is worth investigating that if a term of sequence (2.10 is divisibe by a prime then this prime is a divisor of infinitey many terms of the sequence. The answer is not as obvious as before. First of a we prove a Lemma which hep us in this case and other simiar cases, too. Lemma 2.4. Let a,b be natura numbers with (a,b = 1 and a > 1. If q is a prime divisor of sequence (1.1 then 1. There exists an exponent h q of a with respect to q. 2. If q is a divisor of a k + b then q is a divisor of those terms of sequence (1.1 which can be given of the form where z Z and k + zh p 0. a k+zhp + b Proof. 1. The first statement is trivia. If (a,b = 1 and q is a divisor of a term of sequence (1.1 then (a,q = 1. 2. Let q is a prime divisor of a k +b. Let us denote by h q an exponent of a with respect to q. Let us consider a term in the form a m + b of sequence (1.1. In this case q is a divisor of a m + b if and ony if Using eementary conversions we have (a k + b (a m + b 0 (mod q. (2.15 (a k + b (a m + b = a min{k,m}( a m k 1. Since (a,q = 1 and h q is an exponent of a we get that congruence (2.15 is vaid if and ony if h q is a divisor of m k. This statement is equivaent to our statement. Concusion 2.5. If a prime q is a divisor of two different terms of sequence (2.10 then it is a divisor of infinitey many terms of the sequence.
52 F. Fiip, K. Liptai, J. T. Tóth Proof. Let q be a prime divisor of at east two different terms of sequence (2.10. Let us denote these terms by a p1 + b and a p2 + b where p 1 < p 2. It foows from Lemma 1 that p 2 = p 1 + nh q for some natura number n. Since p 1 and p 2 are primes therefore (p 1,h q = 1. Using Dirichet s theorem we have that there is a subsequence (p n n=1 with prime terms of the sequence (p 1 + nh q n=1. It foows from Lemma 1 that q is a divisor of a terms of the sequence (a p n + b n=1. Further we study a subsequence of (1.1 where the powers are the vaues of Euer s function ϕ. Simiary to the previous sequence the asymptotic density of the set of vaues of Euer s function ϕ equas zero. First we prove that there are infinitey many prime divisors of this sequence. Theorem 2.6. Let a,b be natura numbers where (a,b = 1 and a > 1. Then there are infinitey prime divisors of the sequence (a ϕ(n + b n=1. (2.16 Proof. Let us suppose that there are ony finitey many prime divisors of sequence (2.16 namey q 1,q 2,...,q k. We distinguish two cases. In the first case we suppose that among the prime divisors of sequence (2.16 there are divisors which divide b + 1. Let us denote these divisors by q 1 < < q and the others by q +1 < < q k. Let us denote by α s for a s (1 s the east natura number which Put q αs s > b + 1. M = q α1 1 qα2 2 qα q +1 q k. Obviousy (a,m = 1. It foows from Euer s theorem that a nϕ(m 1 (mod M (2.17 for a natura numbers n. Let us consider an increasing sequence of prime numbers (p i i=1 where (p i,m = 1 for a i N. Since the Euer s function is mutipicative we have that the sequence (a ϕ(piϕ(m + b i=1 (2.18 is a subsequence of sequence (2.16. It is obvious that the prime divisors of sequence (2.18 beong to the set {q 1,q 2,...,q k }.
On prime divisors of remarkabe sequences 53 We choose one of them and et us denote it by q. It foows from (2.17 that 0 a ϕ(piϕ(m + b b + 1 (mod q, that is q is a divisor of b + 1 and q {q 1,q 2,...,q }. Thus we have a ϕ(piϕ(m + b = q βi,1 1 q βi,2 2 q β i, where β i,j 0 for a 1 j and i N. Henceforth we show that β i,j < α j for a 1 j and i N. If β i,j α j for any 1 j and for i N, then we have a ϕ(piϕ(m + b 0 (mod q αj j and aϕ(piϕ(m 1 0 (mod q αj j. It foows from the previous congruences that q αj j the fact that q αj j > b + 1. Hence a ϕ(piϕ(m + b < q αi,1 1 q αi,2 2 q α i, M, which is a contradiction since sequence (2.18 is not bounded. is a divisor of b+1, this contradicts In the second case we suppose that the divisors of sequence (2.16 are not divisors of b + 1. Put L = q 1 q 2 q k. Since (a,l = 1, it foows from the Euer s theorem that a ϕ(l 1 0 (mod L. (2.19 Obviousy a ϕ(l + b is a term of sequence (2.16. Let q be a prime divisor of sequence (2.16. In this case 0 a ϕ(l + b a ϕ(l 1 + b + 1 b + 1 (mod q, that is q is a divisor of b + 1 which is contradiction. Further we investigate when a prime divisor of sequence (2.16 divides infinitey many terms of sequence (2.16. This probem is more difficut than in case (2.10. We give two sufficient conditions. Theorem 2.7. If q is a prime divisor of sequence (2.16 and b + 1 0 (mod q, then q is a divisor of infinitey many terms of sequence (2.16. Proof. Let q be an odd prime divisor of sequence (2.16 with the condition b+1 0 (mod q. Since (a,q = 1, it foows from the Euer s theorem that a ϕ(q 1 (mod q. Obviousy we have a ϕ(q + b a ϕ(q 1 + b + 1 0 (mod q.
54 F. Fiip, K. Liptai, J. T. Tóth Let (p n n=1 be an arbitrary increasing sequence of prime numbers, where q is not a term of this sequence. We show that q is a divisor of a terms of the sequence (a ϕ(qpn + b n=1. Since ϕ is a mutipicative function and (q,p n = 1 we have that a ϕ(qpn + b a ϕ(qϕ(pn + b (a ϕ(q ϕ(pn 1 + b + 1 0 (mod q for a natura numbers n. Theorem 2.8. Let q be a prime divisor of sequence (2.16 and the power of a is an odd number (mod q. Then q is a divisor of infinitey many terms of sequence (2.16. Proof. Let q be such a prime divisor of sequence (2.16 that the power of a is odd (mod q. Let us denote by n 0 the east natura number where q is a divisor of a ϕ(n0 + b. Since the power h q of a is odd (mod q from the Dirichet s theorem foows that the sequence (kh q + 2 k=1 (2.20 contains infinitey many prime numbers. Let us choose a subsequence (p n n=1 of sequence (2.20 which terms are primes and not divisors of the number n 0. Since ϕ mutipicative we have that a ϕ(n0p n + b = a ϕ(n0ϕ(p n + b = a ϕ(n0(p n 1 + b = = a ϕ(n0(khq+1 + b = a ϕ(n0+ϕ(n0khq + b for a n N. Using Lemma 1 we have that q is a divisor of a terms of the sequence (a ϕ(n0p n + b n=1. Finay we show that there are infinitey many primes which do not divide any term of sequence (2.16. First we prove a more genera theorem. Theorem 2.9. Let a > 1 and b > 1 be natura numbers where b is odd and (a,b = 1. Then there are infinitey many primes p which do not divide any term of sequence (a 2n + b n=1 (2.21
On prime divisors of remarkabe sequences 55 Proof. Let p be an arbitrary prime. Then p is not a divisor of any term of sequence (2.21 if and ony if there is no soution of the quadratic congruence x 2 b (mod p. Using the Jacobi s symbo we have ( b = 1. p Let p be an odd prime number where (b, p = 1. Appying the aw of quadratic reciprocity of Gauss we have ( b = p ( 1 p ( p ( = ( 1 p p 1 p 2 b ( 1 p 1 b 1 2 2 = ( p b( 1 p 1 b+1 2 2. (2.22 We distinguish two cases. First we suppose that b = 4 + 1 where is a natura number. Let us consider primes of the form p = 4bk + 2b + 1, where k N. It foows from the Dirichet s theorem that there are infinitey many primes of the form as above since (4b, 2b + 1 = 1. In this case ( ( p b = 1 b = 1 and p 1 b+1 2 2 is odd natura number. Using (2.22 we have ( b = 1. p That is p doesn t divide any term of sequence (2.21. In the second case we suppose that b = 4 + 3 where natura number. Let us consider primes of the form p = 2bk + 2b 1. Using the previous method we get that there are infinitey many primes of this form. Obviousy b+1 2 is even. Moreover ( p ( 1 = = ( 1 b b 1 2 = 1. b Using (2.22 we have equation ( b = 1. p That is p doesn t divide any term of sequence a (2.21. Concusion 2.10. There are infinitey many primes which do not divide any term of sequence (2.16. Proof. Using the previous theorem we get this statement since the Euer s function ϕ is even except those cases when ϕ(1 = ϕ(2 = 1.
56 F. Fiip, K. Liptai, J. T. Tóth References [1] Hardy, G. H., Wright, E. M., An introduction to the theory of numbers, Oxford, 1954. [2] Sárközi, A., Számeméet, Műszaki Könyvkiadó, Budapest, 1976. [3] Sárközi, A., Surányi, J., Számeméet feadatgyűjtemény, 13. kiadás, Tankönyvkiadó, Budapest, 1990. [4] Sierpinsky, W., Eementary theory of numbers, PWN, Warszawa, 1964. [5] Sierpinsky, W., 200 feadat az eemi számeméetbő, Tankönyvkiadó, Budapest, 1964. [6] Tóth, J., Egy számsorozat prímosztóiró, Poygon, Szeged III (2 (1993, 78 79. Ferdinánd Fiip Department of Mathematics University of J. Seye SK-94501 Komárno Rožníckej Škoy 1514 Sovakia Kámán Liptai Institute of Mathematics and Informatics Eszterházy Károy Coege H-3300 Eger Leányka út 4. Hungary János T. Tóth Department of Mathematics University of Ostrava CZ-701 03 Ostrava 30. dubna 22 Czech Repubic