Stewart reduced the problem of determining all Lucas and Lehmer sequences
whose $n$-th element does not have a primitive divisor to solving certain Thue
equations. Using the method of Tzanakis and de Weger for solving Thue
equations, we determine such sequences for $n \leq 30$. Further computations
lead us to conjecture that, for $n > 30$, the $n$-th element of such sequences
always has a primitive divisor

Par Paul M Voutier

Paul M. Voutier

English

arXiv

Paul M Voutier

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Primitive divisors of Lucas and Lehmer sequences, II

Primitive divisors of Lucas and Lehmer sequences, III

Journal de Théorie des Nombres de Bordeaux

Voutier, Paul M.

Numérisation de Documents Anciens Mathématiques

summary:Let $a$, $b$, $c$, $r$ be positive integers such that $a^{2}+b^{2}=c^{r}$, $\min (a,b,c,r)>1$, $\gcd (a,b)=1, a$ is even and $r$ is odd. In this paper we prove that if $b\equiv 3\hspace{4.44443pt}(\@mod \; 4)$ and either $b$ or $c$ is an odd prime power, then the equation $x^{2}+b^{y}=c^{z}$ has only the positive integer solution $(x,y,z)=(a,2,r)$ with $\min (y,z)>1$

Le, Maohua

Institute of Mathematics AS CR, v. v. i.

Czechoslovak Mathematical JournalMaohua LeA note on the diophantine equation x2 + bY = czCzechoslovak Mathematical Journal, Vol. 56 (2006), No. 4, 1109–1116Persistent URL: http://dml.cz/dmlcz/128133Terms of use:© Institute of Mathematics AS CR, 2006Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 56 (131) (2006), 1109–1116A NOTE ON THE DIOPHANTINE EQUATION x2 + bY = czMaohua Le, Guangdong(Received May 19, 2003)Abstract. Let a, b, c, r be positive integers such that a2 + b2 = cr, min(a, b, c, r) > 1,gcd(a, b) = 1, a is even and r is odd. In this paper we prove that if b ≡ 3 (mod 4) andeither b or c is an odd prime power, then the equation x2 + by = cz has only the positiveinteger solution (x, y, z) = (a, 2, r) with min(y, z) > 1.Keywords: exponential diophantine equation, Lucas number, positive divisorMSC 2000 : 11D611. IntroductionLet ,  ,  be the sets of all integers, positive integers and rational numbersrespectively. In 1933, Terai [10] proposed the following conjecture.Conjecture 1. If (a, b, c) is a primitive Pythagorean triple such thata2 + b2 = c2, a, b, c ∈  , gcd(a, b) = 1, a ≡ 0 (mod 2),then the equationx2 + by = cz, x, y, z ∈ has only the solutions (x, y, z) = (a, 2, 2).This problem is related to an early conjecture of Jeśmanowicz [5]. As an analogueof Conjecture 1, Cao and Dong [3] considered the following conjecture:Supported by the National Natural Science Foundation of China (No. 10271104) and theGuangdong Provincial Natural Science Foundation (No. 04011425).1109Conjecture 2. If a, b, c, r, s, t are fixed positive integers such thatax + bt = cr, min(a, b, c, r, s, t) > 1, gcd(a, b) = 1, a ≡ 0 (mod 2),then the equationxs + by = cz, x, y, z ∈ has only the solutions (x, y, z) = (a, t, r).However, the condition min(y, z) > 1 is necessary in Conjuecture 2 (see [4]). Ingeneral, this conjecture is far from solved. In this paper we consider the case that a,b, c, r are fixed positive integers satisfying(1) a2 + b2 = cr, min(a, b, c, r) > 1, gcd(a, b) = 1, a ≡ 0 (mod 2), r 6≡ 0 (mod 2).In this respect, Cao, Dong and Li [4] proved that if(2) a = |Vr|, b = |Ur|, c = m2 + 1and b is an odd prime power with b ≡ 3 (mod 4), where m is an even integer withm > 1 and the integers U(r), V (r) satisfy(3) Vr + Ur√−1 =(m +√−1)r,then the equation(4) x2 + by = cz, x, y, z ∈  min(y, z) > 1has only the solution (x, y, z) = (a, 2, r). In this paper, we show that the condition(2) can be eliminated from the above mentioned result. We shall prove two generalresults:Theorem 1. If (1) holds and b is an odd prime power with b ≡ 3 (mod 4), then(4) has only the solution (x, y, z) = (a, 2, r).Theorem 2. If (1) holds, b ≡ 3 (mod 4) and c is an odd prime power, then (4)has only the solution (x, y, z) = (a, 2, r).11102. Proof of Theorem 1Lemma 1 ([8, pp. 122–123]). Let r be an odd integer with r > 1. Then everysolution (X, Y, Z) of the equationX2 + Y 2 = Zr, X, Y, Z ∈  , gcd(X, Y ) = 1, Y ≡ 0 (mod 2)can be expressed asX + Y√−1 = λ1(m + λ2l√−1)r, λ1, λ2 ∈ {−1, 1},Z = m2 + l2, m, l ∈  , gcd(m, l) = 1, m ≡ 0 (mod 2).Lemma 2. Let k be an odd integer with k > 1, and let ω(k) denote the numberof distinct prime divisors of k. If the equation(5) m2 + l2 = k, m, l ∈  , gcd(m, l) = 1, m ≡ 0 (mod 2)has solutions (m, l), then (5) has exactly 2ω(k)−1 solutions (m, l).	. This lemma follows directly from Lemma 1 of [7]. Lemma 3 ([6]). The equationx2 − 1 = Y n, X, Y, n ∈  , min(X, Y, n) > 1has only the solution (X, Y, n) = (3, 2, 3).Lemma 4 ([9]). Let d is a positive square free integer with square free, and leth(−d) denote the class number of the imaginary quadratic field Q(√−d). If d > 2,then the equation1 + dX2 = Y n, X, Y, n ∈  , Y 6≡ 0 (mod 2),n > 1, n 6≡ 0 (mod 2), h(−d) 6≡ 0 (mod n)has no solutions (X, Y, n).1111Lemma 5. Let p be an odd integer with p ≡ 3 (mod 4). The equation(6) 1 + 3X2 = p2n, X, n ∈  , n 6≡ 0 (mod 2)has only the solution (p, X, n) = (7, 4, 1).	. Since h(−3) = 1, by Lemma 4 we can suppose that n = 1 in (6). Then(u, v) = (p, X) is a solution of the equation(7) u2 − 3v2 = 1, u, v ∈  .Since X is even and 2 +√3 is the fundamental solution of (7), we get(8) p + X√3 =(2 +√3)2t =(7 + 4√3)t, t ∈  ,whence we obtain(9) p =[n/2]∑j=0(t2j)7t−2j48i.Since p ≡ 3 (mod 4), we see from (9) that t is odd. Hence, by (9), we get t = 1 andp = 7. Thus, (6) has only the solution (p, X, n) = (7, 4, 1). The lemma is proved. Lemma 6 ([3, Lemma 1]). Let b be an odd prime power, and let c be a positiveinteger with gcd(b, c) = 1. If (4) has a solution (x, y, z) such that both y and z areeven, then we have(i) b = 239, c = 13, (x, y, z) = (28560, 2, 8).(ii) b2 + 1 = 2c2, (x, y, z) = ( 12 (b2 − 1), 2, 4).(iii) b2t + 1 = 2c, (x, y, z) = ( 12 (b2t − 1), 2t, 4), where t is a positive integer.Let α, β be algebraic integers. If α + β and αβ are nonzero coprime integers andα/β is not a root of unity, then (α, β) is called a Lucas pair. Further, let A = α + βand C = αβ. Then we haveα =12(A + λ√B), β =12(A− λ√B), λ ∈ {−1, 1},where B = A2 − 4C. The numbers of the pair (A, B) are called the parametersof the Lucas pair (α, β). Two Lucas pairs (α1, β1) and (α2, β2) are equivalent ifα1/α2 = β1/β2 = ±1. Given a Lucas pair (α, β), one defines the correspondingsequence of Lucas numbers byLn = Ln(α, β) =αn − βnα− β , n = 0, 1, 2 . . . .1112For equivalent Lucas pairs (α1, β1) and (α2, β2), we have Ln(α1, β1) = ±Ln(α2, β2)for any n > 0. A Prime p is called a primitive divisor of Lt(α, β) if p | Ln andBL1 . . . Ln−1 6≡ 0 (mod p). A Lucas pair (α, β) such that Ln(α, β) has no primitivedivisors will be called an n-defective Lucas pair. Further, a positive integer n iscalled totally non-defective if no Lucas pair is n-defective.Lemma 7 ([11]). Let n satisfy 4 < n 6 30 and n 6= 6. Then, up to equivalence,all parameters of n-defective Lucas pairs are given as follows:(i) n = 5, (A, B) = (1, 5), (1,−7), (2,−40), (1,−11), (1,−15), (12,−76), (12,−1364).(ii) n = 7, (A, B) = (1,−7), (1,−19).(iii) n = 8, (A, B) = (2,−24), (1,−7).(iv) n = 10, (A, B) = (2,−8), (5,−3), (5,−47).(v) n = 12, (A, B) = (1, 5), (1,−7), (1,−11), (2,−56), (1,−15), (1,−19).(vi) n ∈ {13, 18, 30}, (A, B) = (1,−7).Lemma 8 ([1, Theorem 1.4]). If n > 30, then n is totally non-defective.Lemma 9. If a, b, c, r satisfy (1) and b is an odd prime with b ≡ 3 (mod 4), theneither (a, b, c, r) = (524, 7, 65, 3) or a, b, c and r satisfy (2).	. By Lemma 1, we get from (1) thata + b√−1 = λ1(m + λ2l√−1)r, λ1, λ2 ∈ {−1, 1},(10)c = m2 + l2, m, l ∈  , gcd(m, l) = 1, m ≡ 0 (mod 2).(11)From (10), we obtain(12) b = λ1λ2l(r−1)/2∑i=0(r2i + 1)mm−2i−1(−l2)i.Since b is an odd prime power with b ≡ 3 (mod 4), we have(13) b = pk,where p is an odd prime and k is an odd integer. By (12) and (13), we get(14) l = ps,∣∣∣∣(r−1)/2∑i=0(r2i + 1)mr−2i−1(−l2)i∣∣∣∣ = pk−s, s ∈ , 0 6 s 6 k.By (3), (10), (11) and (14), if s = 0, then a, b, c, r satisfy (2). If s > 0, let(15) α = m + l√−1, β = m− l√−1.1113Then (α, β) is a Lucas pair with parameters (2m,−4l2). Further, let Ln(α, β) (n > 0)denote the corresponding Lucas numbers. Then, by (14), we get(16) l = ps, |Lr(α, β)| = pk−s, 0 < s 6 k.It implies that the Lucas number Lr(α, β) has no primtitive divisors. Since r is anodd integer with r > 1, by Lemmas 7 and 8 we obtain r = 3.When r = 3 and s = k, we get from (14) that(17) p2s − 3m2 = 1.Since b ≡ 3 (mod 4), we see from (13) that p ≡ 3 (mod 4). Hence, by Lemma 5, weget from (17) that p = 7, s = 1 and m = 4. Therefore, by (10) and (11), we abtain(a, b, c, r) = (524, 7, 65, 3).When r = 3 and s < k, since s > 0 and gcd(m, l) = 1, we get from (14) thatp = 3, k − s = 1 and(18) m2 − 32s−1 = 1.By Lemma 3, we find from (18) that m = 2 and s = 1. Hence, by (13), we getb = 32 = 9. But, since b ≡ 3 (mod 4), this is impossible. Thus the lemma is proved.	of Theorem 1. Since b ≡ 3 (mod 4), by Theorem of [4] and our Lemma9 it suffices to prove the theorem for (a, b, c, r) = (524, 7, 65, 3). Then (4) can bewritten as(19) x2 + 7y = 65z, x, y, z ∈  , min(y, z) > 1.Let (x, y, z) be a solution of (19) with (x, y, z) 6= (524, 2, 3). By Lemma 6, we havey ≡ 0 (mod 2) and z 6≡ 0 (mod 2). Hence, by Lemma 1, we getx + 7y/2√−1 = λ1(m + λ2l√−1)z, λ1, λ2 ∈ {−1, 1},(20)65 = m2 + l2, m, l ∈  , gcd(m, l) = 1, m ≡ 0 (mod 2).(21)Since ω(65) = 2, by Lemma (2), (21) has exactly two solutions (m, l) = (4, 7) and(8,1).When (m, l) = (4, 7), let(22) α = 4 + 7√−1, β = 4− 7√−1.1114Then (α, β) is a Lucas pair with parameters (8, 196). Further, let Ln(α, β) (n > 0)denote the corresponding Lucas numbers. Then, from (20) and (22) we get(23) 7y/2−1 = |Lz(α, β)|.This implies that the Lucas number Lz(α, β) has no primtitive divisors. On the otherhand, since z > 1 and (x, y, z) 6= (524, 2, 3), we see from (20) that z > 3. But, byLemmas 7 and 8, (23) is impossible.When (m, l) = (8, 1), we get from (20) that(24) 7y/2 = λ1λ2(z−1)/2∑i=0(−1)i(z2i + 1)8z−2i−1.Since 8 ≡ 1 (mod 7) and(z−1)/2∑i=0(−1)i(z2i + 1)=z∑j=0(zj)sinjπ2(25)= 2z sinzπ4(cosπ4)z= (−1)(z−1)(z+5)/82(z−1)/2,by (24), we obtain 0 ≡ ±2(z−1)/2 (mod 7), a contradiction. Thus, (4) has only thesolution (x, y, z) = (524, 2, 3) for (a, b, c, r) = (524, 7, 65, 3). The theorem is proved.3. Proof of Theorem 2Lemma 10 ([2, Theorem 4]). Let D be a positive integer with D > 2, and let pbe an odd prime with D 6≡ 0 (mod p). If (D, p) = (3s2 + 1, 4s2 + 1), where s is apositive integer, then the equation(26) X2 + DY = pz, X, Y, Z ∈ has at most three solutions (X, Y, Z) = (s, 1, 1), (8s2 + 3s, 1, 3) and (X3, Y3, Z3),where Y3 is even. Otherwise, (26) has at most two solutions (X, Y, Z). Further, ifthese are (X1, Y1, Z1) and (X2, Y2, Z2), then Y1 ≡ Y2 (mod 2).	of Theorem 2. Since c is an odd prime power, we have c = pt, wherep is an odd prime and t is a positive integer. Hence, if (x, y, z) is a solution of (4),then (X, Y, Z) = (x, y, tz), is a solution of the equation(27) X2 + bY = pZ , X, Y, Z ∈  .1115Since b ≡ 3 (mod 4), hence if (4) has a solution (x, y, z) 6= (a, 2, r), then (27) hasat least two solutions (X, Y, Z) with Y ≡ 0 (mod 2). But, by Lemma 10, this isimpossible. Thus, the theorem is proved. References[1] Y.Bilu, G.Hanrot and P.Voutier (with an appendix by M.Mignotte): Existence of prim-itive divisors of Lucas and Lehmer numbers. J. Reine Angew. Math. 539 (2001), 75–122.[2] Y.Bugeaud: On some exponential diophantine equations. Monatsh. Math. 132 (2001),93–97.[3] Z.-F.Cao and X.-L.Dong: The diophantine equation a2 + by = cz. Proc. Japan Acad.77A (2001), 1–4.[4] Z.-F.Cao, X.-L.Dong and Z. Li: A new conjecture concerning the diophantine equationx2 + by = cz . Proc. Japan Acad. 78A (2002), 199–202.[5] L. Jeśmanowicz: Several remarks on Pythagorean number.Wiadom. Mat. 1 (1955/1956),196–202. (In Polish.)[6] C.Ko: On the diophantine equation x2 = yn + 1, xy 6= 0. Sci.Sin. 14 (1964), 457–460.[7] M.-H.Le: A note on Jeśmanowicz’ conjecture. Colloq. Math. 64 (1995), 47–51.[8] L. J.Mordell: Diophantine Equations. Academic Press, London, 1969.[9] T.Nagell: Sur I’impossibilité de quelques equation á deux indéterminées. Norsk Matem.Forenings Skrifter 13 (1921), 65–82.[10] N.Terai: The diophantine equation x2 + qm = pn. Acta Arith. 63 (1993), 351–358.[11] P.Voutier: Primitive divisors of Lucas and Lehmer sequences. Math. Comp. 64 (1995),869–888.Author’s address: Department of Mathematics, Zhanjiang Normal College, Zhanjiang,Guangdong, P. R.China.1116

A note on the diophantine equation $x^2+b^Y=c^z$

Let ff and fi are conjugate complex algebraic integers which generate Lucas or Lehmer sequences. We present an algorithm to search for elements of such sequences which have no primitive divisors. We use this algorithm to prove that for all ff and fi with h(fi=ff)  4, the n-th element of these sequences has a primitive divisor for n ? 30. In the course of proving this result, we give an improvement of a result of Stewart concerning more general sequences. 1. Introduction  Let ff and fi be algebraic numbers such that ff + fi and fffi are relatively prime non-zero rational integers and ff=fi is not a root of unity. The sequence (u n )  1  n=0  defined by u n =  ff  n  \Gamma fi  n  ff \Gamma fi  for n  0 is called a Lucas sequence.  If, instead of supposing that ff + fi 2 Z, we only suppose that (ff + fi)  2  is a non-zero rational integer, still relatively prime to fffi, then we define the  Lehmer sequence (u n )  1  n=0 associated to ff and fi by u n =  8 ? ? ! ? ? :  ff  n  \Gamma fi  ..

Stewart reduced the problem of determining all Lucas and Lehmer sequences whose nth element does not have a primitive divisor to solving certain Thue equations. Using the method of Tzanakis and de Weger for solving Thue equations, we determine such sequences for 

  
    
      n
      ≤
      30
    
    n \leq 30
  

. Further computations lead us to conjecture that, for 

  
    
      n
      &gt;
      30
    
    n &gt; 30
  

, the nth element of such sequences always has a primitive divisor.</p

Primitive divisors of Lucas and Lehmer sequences

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