Filip, Ferdinánd

Liptai, Kálmán

Tóth T., János

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Annales Mathematicae et Informaticae33 (2006) pp. 45–56http://www.ektf.hu/tanszek/matematika/amiOn prime divisors of remarkable sequencesFerdinánd Filipa, Kálmán Liptaib1, János T. Tóthc2aDepartment of Mathematics University of J. Selyee-mail: ﬁlip.ferdinand@seznam.czbInstitute of Mathematics and Informatics Eszterházy Károly Collegee-mail: liptaik@ektf.hucDepartment of Mathematics University of Ostravae-mail: toth@osu.czSubmitted 10 November 2006; Accepted 18 December 2006AbstractIn this paper we study sequences of the form (an + b)∞n=1, where a, b ∈ N.We prove many interesting results connection with sequences with inﬁnitelymany prime divisors.Keywords: prime divisors, Dirichlet’s theoremMSC: 11N131. IntroductionThere are many mathematical problems when we investigate the divisibility ofsequences by a prime. We usually find this kind of interesting examples in nationalmathematical competitions and in the International Math Olympiad. In this paperwe study sequences of the form (an + b)∞n=1, where a, b ∈ N. We prove someresults concerning with sequences with infinitely many prime divisors. Moreover wecharacterize these sequences. Some of our theorems assert that there are infinitelymany prime divisors of a sequence. These statements come from easily from thetheory of S-units, but in this paper we use only elementary methods to get ourresults. We mention that our results help to generalize problems which can befound in some exercise books for students.1Research supported by the Hungarian National Foundation for Scientiﬁc Research Grant. No.T 048945 MAT2Research supported by Grant ČR 201/04/0381/24546 F. Filip, K. Liptai, J. T. TóthLet A = {a1 < a2 < · · · < an < · · · } ⊆ N be a given set and let us denoteby A(x) the number of the elements of A not exceeding x. Let us suppose for anynatural number k there is a positive real number xk such that for all x > xk theinequality A(x) > (log x)k holds. In this case there are infinitely many differentprime divisors of the elements of A (see [3], p. 102).Further we shall study the sequences of positive integers where the previouscondition is not true. Let a, b be natural numbers with a > 1 and (a, b) = 1.Obviously the sequences(an + b)∞n=1 (1.1)do not fulfill the above condition, sinceA(x) =[log(x− b)log a]if x > b+ 1.In what follows we show that sequences (1.1) have infinitely many differentprime divisors. In the special case, when a = 10 and b = 3 we proved (in [6])that the sequence (10n + 3)∞n=1 has infinitely many prime divisors, moreover forinfinitely many primes p there are infinitely many n ∈ N such that p | 10n + 3.2. ResultsFirst we prove that there are subsequences of sequences (1.1) which have infi-nitely many prime divisors.Theorem 2.1. Let a, b, c, d be natural numbers, (a, b) = 1 and a > 1. Then thereare infinitely many prime divisors of the sequences(ac+(n−1)d + b)∞n=1. (2.1)Proof. First we suppose that sequence (2.1) has only finitely many prime divisors.Let us denote these primes by q1 < q2 < · · · < qk. Let us denote by q1 < q2 <· · · < ql the prime divisors of sequence (2.1) which are divisors of ac+ b as well andql+1 < ql+2 < · · · < qk which are not divisors of ac + b. Let us denote by αs for all1 6 s 6 l and s ∈ N the least natural number such thatqαss > ac + b.LetM = qα11 qα22 · · · qαll ql+1ql+2 · · · qkbe a product of prime powers. In this case (a,M) = 1 since (a, b) = 1. By thetheorem of Euler we haveM | anϕ(M) − 1 (2.2)for all n ∈ N.On prime divisors of remarkable sequences 47Now we investigate the sequence(ac+mϕ(M)d + b)∞m=1 (2.3)which is obviously a subsequence of sequence (2.1).Let q be a prime divisor of sequence (2.3) that isamϕ(M)d+c + b ≡ 0 (mod q) (2.4)for some m ∈ N. It follows from (2.2) thatamϕ(M)d − 1 ≡ 0 (mod q). (2.5)Using (2.4) and (2.5) we haveamϕ(M)d+c + b = amϕ(M)d(ac − 1) + amϕ(M)d − 1 + b+ 1 ≡ ac + b (mod q).It is clear that q | ac + b, it follows that q ∈ {q1, q2, . . . , ql}, that isamϕ(M)d+c + b = qβm11 qβm22 · · · qβmllwhere βmj > 0 for all m ∈ N and 1 6 j 6 l.We show that for all m ∈ N and 1 6 j 6 l we have βmj < αj . Let 1 6 j 6 l, mbe arbitrary natural numbers and βmj > αj thenqαjj | amϕ(M)d+c + b,that isamϕ(M)d+c + b ≡ 0 (mod qαjj ).Since qαjj |M , it follows from (2.2) that amϕ(M)d − 1 ≡ 0 (mod qαjj ) andamϕ(M)d+c + b = amϕ(M)d(ac − 1) + amϕ(M)d − 1 + b+ 1 ≡ ac + b (mod qαjj ),that is qαjj | ac + b, which is contradiction since qαjj > ac + b. It follows that for allterms of (2.3) we haveamϕ(M)d+c + b < qα11 qα22 · · · qαll 6M .In this way we obtained a contradiction since sequence (2.3) is not bounded. In the sequel 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 dividesinfinitely many terms of sequence (2.1).48 F. Filip, K. Liptai, J. T. TóthProof. Let m ∈ N be a divisor of a term of sequence (2.1). Let us denote by n0the least non-negative number whichm | ac+n0d + b . (2.6)Since (a,m) = 1, there exists a power hm of a (mod m). The number m dividesan − 1 if and only if hm | n.Let us consider the sequence(ankd+c + b)∞n=1 (2.7)wherenk = (k − 1)hm(hm, d)+ n0 .Obviously sequence (2.7) is a subsequence of sequence (2.1).We show that m divides only those terms of sequence (2.1) which are the termsof (2.7) as well.a) First we prove that m divides all terms of sequence (2.7). Obviously we haveankd+c + b = ankd+c + b− an0d+c + an0d+c == an0d+c(a(nk−n0)d − 1) + ank0+c + b == an0d+c(a(k−1) hmd(hm,d) − 1)+ an0d+c + b.(2.8)Using that d(hm,d) is an integer number and the definition of hm we havea(k−1) d(hm,d)hm − 1 ≡ 0 (mod m).It follows thatankd+c + b ≡ an0d+c + b (mod m),that is m divides all terms of (2.7).b) Secondly we prove that if m divides a term of sequence (2.1) then this termis a term of sequence (2.7).Let us choose n ∈ N such that m | and+c1 + b. Obviously n > n0. Then we havem | and+c1 + b−(an0d+c1 + b)= an0d+c1(ad(n−n0) − 1).Since (a,m) = 1, therefore m | ad(n−n0) − 1. Using the definition of hm we havehm | d(n− n0), andn = (k − 1)hmd+ n0 (2.9)for some k ∈ N. From equation (2.9) we deducen =((k − 1)hm(hm,d)d(hm,d))+ n0 .On prime divisors of remarkable sequences 49Using that(hm(hm,d), d(hm,d))= 1, we have that n is an integer if and only if thefraction k−1d(hm,d)is also an integer. Consequentlyk − 1 = (l − 1)d(hm, d),andn = (l − 1)hm(hm, d)+ n0.Now the theorem is proved. In the previous theorems we investigated such subsequences of sequences (1.1)where the powers formed arithmetic progressions. It is known that the asymptoticdensity of sets of terms of arithmetic progressions are greater than zero, moreexactly it equals the reciprocal of the difference. This means that sequence (2.1)is such a subsequence of (1.1) which contains relatively “many” terms of sequence(1.1). In what follows we are looking for subsequences of (1.1) where the density ofthe set of powers is zero, but they have infinitely many prime divisors. We give twosequences possessing the above conditions. In one of them the powers run throughthe set of primes and in the other the powers equal the values of Euler’s functionϕ. It is known fact that the asymptotic density of the set of primes and the set ofvalues of Euler’s function are zero.Theorem 2.3. Let a, b be natural numbers with (a, b) = 1 and a > 1. Let usdenote by pn the n-th prime number. Then the sequence(apn + b)∞n=1 (2.10)has infinitely many prime divisors.Proof. Let us suppose that sequence (2.10) has only finitely many prime divisors,namely q1, q2, . . . , qk. 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 q1 < · · · < ql the divisors of a + b andql+1 < · · · < qk which are not divisors of a + b. Let us denote by αs for all1 6 s 6 l the least natural number whichqαss > a+ b .PutM = qα11 qα22 · · · qαll ql+1 · · · qk.In this case (a,M) = 1 since (a, b) = 1. It follows from Euler’s theorem thatanϕ(M) − 1 ≡ 0 (mod M) (2.11)50 F. Filip, K. Liptai, J. T. Tóthfor all n ∈ N. Using the theorem of Dirichlet we get that there are infinitely manyprime numbers in the sequence (nϕ(M)+1)∞n=1. Let us denote these prime numbersby p′1 < p′2 < · · · p′n < · · · . Obviously the sequence(ap′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). Obviously q ∈ {q1, q2, . . . , qk}, moreoverap′i + b ≡ 0 (mod q) (2.13)for some i ∈ N. It follows from (2.11) and (2.13) that0 ≡ ap′i + b ≡ ap′i−1(a− 1) + ap′i−1 + b ≡ a+ b (mod q).Thus q | a+ b and q ∈ {q1, q2, . . . , ql}. In other words ap′i + b can be written in theformap′i + b = qβi,11 qβi,22 · · · qβi,llwhere βi,j > 0 for all 1 6 j 6 l natural numbers.Now we show that βi,j < αj for all 1 6 j 6 l. If βi,j > αj for some 1 6 j 6 lthenap′i + b ≡ 0 (mod qαjj )moreover using (2.11) and qαjj |M we haveap′i−1 ≡ 1 (mod qαjj ).It follows from the previous congruence that0 ≡ ap′i + b ≡ ap′i−1(a− 1) + ap′i−1 + b ≡ a+ b (mod qαjj )which contradicts the fact that qαjj > a+ b. In this way we getap′i + b < qα11 qα22 · · · qαll 6Mfor all i ∈ N. Here we have obtained a contradiction since sequence (2.12) is notbounded.In the second case we study when the terms of sequence (2.10) do not have suchprime divisors which divide a+ b. PutL = q1q2 · · · qk.Since (a, L) = 1, thereforeanϕ(L) − 1 ≡ 0 (mod L) (2.14)for all n ∈ N. LetQ = lϕ(L) + 1On prime divisors of remarkable sequences 51be a prime and q be a prime divisor of aQ + b.It follows from the definition of Q and from (2.14) thataQ−1 ≡ 1 (mod q)where q ∈ {q1, q2, . . . , qk}. Obviously0 ≡ aQ + b ≡ aQ−1(a− 1) + aQ−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 divisible by a primethen this prime is a divisor of infinitely many terms of the sequence. The answeris not as obvious as before. First of all we prove a Lemma which help us in thiscase and other similar cases, too.Lemma 2.4. Let a, b be natural numbers with (a, b) = 1 and a > 1. If q is a primedivisor of sequence (1.1) then1. There exists an exponent hq of a with respect to q.2. If q is a divisor of ak + b then q is a divisor of those terms of sequence (1.1)which can be given of the formak+zhp + bwhere z ∈ Z and k + zhp > 0.Proof. 1. The first statement is trivial. If (a, b) = 1 and q is a divisor of a termof sequence (1.1) then (a, q) = 1.2. Let q is a prime divisor of ak + b. Let us denote by hq an exponent of a withrespect to q. Let us consider a term in the form am + b of sequence (1.1). In thiscase q is a divisor of am + b if and only if(ak + b)− (am + b) ≡ 0 (mod q). (2.15)Using elementary conversions we have(ak + b)− (am + b) = amin{k,m}(a|m−k| − 1).Since (a, q) = 1 and hq is an exponent of a we get that congruence (2.15) is valid ifand only if hq is a divisor of |m−k|. This statement is equivalent to our statement.Conclusion 2.5. If a prime q is a divisor of two different terms of sequence (2.10)then it is a divisor of infinitely many terms of the sequence.52 F. Filip, K. Liptai, J. T. TóthProof. Let q be a prime divisor of at least two different terms of sequence (2.10).Let us denote these terms by ap1 + b and ap2 + b where p1 < p2. It follows fromLemma 1 thatp2 = p1 + nhqfor some natural number n. Since p1 and p2 are primes therefore (p1, hq) = 1.Using Dirichlet’s theorem we have that there is a subsequence (p′n)∞n=1 with primeterms of the sequence (p1 + nhq)∞n=1. It follows from Lemma 1 that q is a divisorof all terms of the sequence(ap′n + b)∞n=1.Further we study a subsequence of (1.1) where the powers are the values ofEuler’s function ϕ. Similarly to the previous sequence the asymptotic density ofthe set of values of Euler’s function ϕ equals zero. First we prove that there areinfinitely many prime divisors of this sequence.Theorem 2.6. Let a, b be natural numbers where (a, b) = 1 and a > 1. Then thereare infinitely prime divisors of the sequence(aϕ(n) + b)∞n=1. (2.16)Proof. Let us suppose that there are only finitely many prime divisors of sequence(2.16) namely q1, q2, . . . , qk. 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 q1 < · · · < qland the others by ql+1 < · · · < qk.Let us denote by αs for all s (1 6 s 6 l) the least natural number whichqαss > b+ 1 .PutM = qα11 qα22 · · · qαll ql+1 · · · qk.Obviously (a,M) = 1. It follows from Euler’s theorem thatanϕ(M) ≡ 1 (mod M) (2.17)for all natural numbers n. Let us consider an increasing sequence of prime numbers(pi)∞i=1 where (pi,M) = 1 for all i ∈ N. Since the Euler’s function is multiplicativewe have that the sequence(aϕ(pi)ϕ(M) + b)∞i=1 (2.18)is a subsequence of sequence (2.16).It is obvious that the prime divisors of sequence (2.18) belong to the set{q1, q2, . . . , qk}.On prime divisors of remarkable sequences 53We choose one of them and let us denote it by q. It follows from (2.17) that0 ≡ aϕ(pi)ϕ(M) + b ≡ b+ 1 (mod q),that is q is a divisor of b+ 1 and q ∈ {q1, q2, . . . , ql}. Thus we haveaϕ(pi)ϕ(M) + b = qβi,11 qβi,22 · · · qβi,llwhere βi,j > 0 for all 1 6 j 6 l and i ∈ N.Henceforth we show that βi,j < αj for all 1 6 j 6 l and i ∈ N. If βi,j > αj forany 1 6 j 6 l and for i ∈ N , then we haveaϕ(pi)ϕ(M) + b ≡ 0 (mod qαjj ) and aϕ(pi)ϕ(M) − 1 ≡ 0 (mod qαjj ).It follows from the previous congruences that qαjj is a divisor of b+1, this contradictsthe fact that qαjj > b+ 1. Henceaϕ(pi)ϕ(M) + b < qαi,11 qαi,22 · · · qαi,ll 6M ,which is a contradiction since sequence (2.18) is not bounded.In the second case we suppose that the divisors of sequence (2.16) are notdivisors of b+ 1. PutL = q1q2 · · · qk.Since (a, L) = 1, it follows from the Euler’s theorem thataϕ(L) − 1 ≡ 0 (mod L). (2.19)Obviously aϕ(L) + b is a term of sequence (2.16). Let q be a prime divisor ofsequence (2.16). In this case0 ≡ aϕ(L) + b ≡ aϕ(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 infinitelymany terms of sequence (2.16). This problem is more difficult 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 infinitely 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 follows from the Euler’s theorem that aϕ(q) ≡ 1(mod q). Obviously we haveaϕ(q) + b ≡ aϕ(q) − 1 + b+ 1 ≡ 0 (mod q).54 F. Filip, K. Liptai, J. T. TóthLet (pn)∞n=1 be an arbitrary increasing sequence of prime numbers, where q is nota term of this sequence.We show that q is a divisor of all terms of the sequence(aϕ(qpn) + b)∞n=1.Since ϕ is a multiplicative function and (q, pn) = 1 we have thataϕ(qpn) + b ≡ aϕ(q)ϕ(pn) + b ≡ (aϕ(q))ϕ(pn) − 1 + b+ 1 ≡ 0 (mod q)for all natural numbers n. Theorem 2.8. Let q be a prime divisor of sequence (2.16) and the power of a isan odd number (mod q). Then q is a divisor of infinitely many terms of sequence(2.16).Proof. Let q be such a prime divisor of sequence (2.16) that the power of a isodd (mod q). Let us denote by n0 the least natural number where q is a divisorof aϕ(n0) + b. Since the power hq of a is odd (mod q) from the Dirichlet’s theoremfollows that the sequence(khq + 2)∞k=1 (2.20)contains infinitely many prime numbers. Let us choose a subsequence(p′n)∞n=1of sequence (2.20) which terms are primes and not divisors of the number n0. Sinceϕ multiplicative we have thataϕ(n0p′n) + b = aϕ(n0)ϕ(p′n) + b = aϕ(n0)(p′n−1) + b == aϕ(n0)(khq+1) + b = aϕ(n0)+ϕ(n0)khq + bfor all n ∈ N. Using Lemma 1 we have that q is a divisor of all terms of the sequence(aϕ(n0p′n) + b)∞n=1.Finally we show that there are infinitely many primes which do not divide anyterm of sequence (2.16). First we prove a more general theorem.Theorem 2.9. Let a > 1 and b > 1 be natural numbers where b is odd and (a, b) =1. Then there are infinitely many primes p which do not divide any term of sequence(a2n + b)∞n=1 (2.21)On prime divisors of remarkable sequences 55Proof. Let p be an arbitrary prime. Then p is not a divisor of any term of sequence(2.21) if and only if there is no solution of the quadratic congruence x2 ≡ −b(mod p). Using the Jacobi’s symbol we have(−bp)= −1.Let p be an odd prime number where (b, p) = 1. Applying the law of quadraticreciprocity of Gauss we have(−bp)=(−1p)(pp)= (−1)p−12(pb)(−1)p−12b−12 =(pb)(−1)p−12b+12 . (2.22)We distinguish two cases.First we suppose that b = 4l + 1 where l is a natural number. Let us considerprimes of the formp = 4bk + 2b+ 1, where k ∈ N.It follows from the Dirichlet’s theorem that there are infinitely many primes of theform as above since (4b, 2b + 1) = 1.In this case(pb)=(1b)= 1 and p−12b+12 is odd natural number. Using (2.22)we have (−bp)= −1.That is p doesn’t divide any term of sequence (2.21).In the second case we suppose that b = 4l + 3 where l natural number.Let us consider primes of the formp = 2bk + 2b− 1.Using the previous method we get that there are infinitely many primes of thisform. Obviously b+12 is even. Moreover(pb)=(−1b)= (−1)b−12 = −1.Using (2.22) we have equation(−bp)= −1.That is p doesn’t divide any term of sequence a (2.21). Conclusion 2.10. There are infinitely many primes which do not divide any termof sequence (2.16).Proof. Using the previous theorem we get this statement since the Euler’s functionϕ is even except those cases when ϕ(1) = ϕ(2) = 1. 56 F. Filip, K. Liptai, J. T. TóthReferences[1] Hardy, G. H., Wright, E. M., An introduction to the theory of numbers, Oxford,1954.[2] Sárközi, A., Számelmélet, Műszaki Könyvkiadó, Budapest, 1976.[3] Sárközi, A., Surányi, J., Számelmélet feladatgyűjtemény, 13. kiadás, Tankönyvki-adó, Budapest, 1990.[4] Sierpinsky, W., Elementary theory of numbers, PWN, Warszawa, 1964.[5] Sierpinsky, W., 200 feladat az elemi számelméletből, Tankönyvkiadó, Budapest,1964.[6] Tóth, J., Egy számsorozat prímosztóiról, Polygon, Szeged III (2) (1993), 78–79.Ferdinánd FilipDepartment of MathematicsUniversity of J. SelyeSK-94501 KomárnoRožníckej Školy 1514SlovakiaKálmán LiptaiInstitute of Mathematics and InformaticsEszterházy Károly CollegeH-3300 EgerLeányka út 4.HungaryJános T. TóthDepartment of MathematicsUniversity of OstravaCZ-701 03 Ostrava30. dubna 22Czech Republic

On prime divisors of remarkable sequences

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On prime divisors of remarkable sequences

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