Diophantine triples in a Lucas-Lehmer sequence

Transcription

1 Annales Mathematicae et Informaticae 49 (01) pp doi: /ami Diophantine triples in a Lucas-Lehmer sequence Krisztián Gueth Lorand Eötvös University Savaria Department of Mathematics Károli Gáspár tér Szombathely Hungary guethk@gmail.com Submitted October 19, 017 Accepted August 4, 01 Abstract In this paper, we define a Lucas-Lehmer type sequence denoted by (L n) n=0, and show that there are no integers 0 < a < b < c such that ab + 1, ac + 1 and bc + 1 all are terms of the sequence. Keywords: Diophantine triples, Lucas-Lehmer sequences MSC: Primary 11B39; Secondary 11D99 1. Introduction A diophantine m-tuple consists of m distinct positive integers such that the product of any two of them is one less than a square of an integer. Diophantus found the first four, but rational numbers 1/16, 33/16, 17/4, 105/16 with this property. Fermat gave 1, 3,, 10 as the first integer quadruple. Hoggatt and Bergum [] provided infinitely many diophantine quadruples by F k, F k+, F k+4, 4F k+1 F k+ F k+3. The most outstanding result is due to Dujella [3], who proved that there are only finitely many quintuples. Recently He, Togbe, and Ziegler submitted a work which solved the longstanding problem of the non-existence of diophantine quintuples [7]. There are several variations of the basic problem, most of them replace the squares by a given infinite set of integers. For instance, Luca and Szalay studied the diophantine triples for the terms of binary recurrences. They proved that there 5

2 6 K. Gueth are no integers 0 < a < b < c such that ab + 1, ac + 1 and bc + 1 all are Fibonacci numbers (see [9]), further for the Lucas sequence there is only one such a triple: a = 1, b =, c = 3 (see [10]). Fuchs, Luca and Szalay [4] gave sufficient and necessary conditions to have infinitly many diophantine triples for a general second order sequence. For ternary recurrences Fuchs et al. [5] justified that there exist only finitely many triples corresponding to Tribonacci sequence. This paper was generalized by Fuchs et al. [6]. Alp and Irmak were the first who investigated the existence of diophantine triples in a Lucas-Lehmer type sequence (see []). They showed that there are no diophantine triples for the so-called pellans sequence. In this paper, we study another Lucas-Lehmer sequence and prove the nonexistence of diophantine triples associated to it. Let (L n ) n=0 be defined by the initial values L 0 = 0, L 1 = 1, L = 1 and L 3 = 3, and by the recursive rule Our principal result is the following. L n = 4L n L n 4. (1.1) Theorem 1.1. There exist no integers 0 < a < b < c such that ab + 1 = L x, ac + 1 = L y, bc + 1 = L z (1.) would hold for any positive integers x, y and z.. Preliminaries The associate sequence of (L n ) is denoted by (M n ) n=0, which according to the general theory of Lucas-Lehmer sequences satisfies M 0 =, M 1 =, M = 4, M 3 = 10, and M n = 4M n M n 4. It is easy to see that L n is divisible by 4 if and only if 4 n, otherwise L n is odd. Using the recurrence relation (1.1), for negative subscripts M n = ( 1) n M n follows. The zeros of the common characteristic polynomial x 4 4x + 1 of (L n ) and (M n ) are ω = ( 3 + 1)/, ψ = ( 3 + 1)/, ω and ψ, further the initial values provide the explicit formulae L n = (ωn ψ n ) (( ω)n ( ψ) n ), M n = 1 + (ω n + ψ n ) + 1 (( ω) n + ( ψ) n ). (.1) It s trivial from the recursive rules of both (L n ) and (M n ) that the subsequences of terms with even resp. odd indices form second order sequences by the same coefficients. The zeros of their companion polynomial are α = ω = + 3 and β = ψ = 3, and the dominant root is α. Generally the Lucas-Lehmer sequences are union of two binary recursive sequences. Many properties, which are well known for binary sequences with initial

3 Diophantine triples in a Lucas-Lehmer sequence 7 values 0 and 1, hold for Lucas-Lehmer sequences too (may be by a little modification). So the research of Lucas-Lehmer sequences is a new feature in the investigations. In the sequel, we prove a few lemma which will be useful in proving the main theorem. Lemma.1. If n = mt and t is odd, then M m M n. Proof. The statement is obvious for t = 1. Formula (.1) admits M 6k = M k (M 4k 1), (.) M 6k+3 = M k+1 (M 4k+ + 1), (.3) which proves the lemma for t = 3. It can be seen by induction on k that M n+k = { 1 M nm k + M n k, if n k 1 (mod ), M n M k ( 1) k M n k, otherwise. (.4) Finally, using (.4), we can prove the lemma by induction on t. Lemma.. If n = mt and t is even, then gcd(m n, M m ) =. Proof. Put m = k. From (.1) it follows that M 4k = M k. (.5) Subsequently, gcd(m k, M 4k ) =. It can be seen that M l k (l 3) can be expressed as a polynomial of M k, where the constant term is always. Thus gcd(m k, M k) = (l ). l Now let m = k + 1. Again by (.1) we see that M 4k+ = M k+1/ + (.6) holds. Putting H k+1 = M k+1 /, it is trivial that H k+1 and M k+1 are divisible by the same primes, and the exponent of is 1 in both integers. So gcd(h k+1, N) = and gcd(m k+1, N) = are equivalent for an arbitrary integer N. Hence we have M 4k+ = H k+1 +, and it implies gcd(m 4k+, H k+1 ) =. By induction and (.5) we can see that M l (k+1) can be written as a polynomial of H k+1 for any positive integer l, with constant term. Consequently, gcd(m k+1, M l (k+1)) = gcd(h k+1, M l (k+1)) =. Together with Lemma.1, it shows immediately, that gcd(m m, M tm ) = for arbitrary even t. Lemma.3. For any n 0 we have L n 1 = L n 1 L n+1 1 L n+ L n M n+1 M n 1 M n M n+, if n 1 (mod 4),, if n 3 (mod 4),, if n 0 (mod 4),, if n (mod 4). (.7)

4 K. Gueth Proof. To prove the statement one can use the explicit formulae for the terms appearing in (.7). Lemma.4. The greatest common divisors of the terms of (L n ) and (M n ) satisfy 1. gcd(l m, L n ) = L gcd(m,n) ; { m Mgcd(m,n), if. gcd(m m, M n ) = gcd(m,n) 1, otherwise; { m µmgcd(m,n), if 3. gcd(l m, M n ) = gcd(m,n) or, otherwise, where µ = 1 or 1/. n gcd(m,n) (mod ), n gcd(m,n) (mod ), Proof. We omit the proof of the first statement, the easiest part, and start by proving the second one. The main tool is a Euclidean-like algorithm. Assume that m = nq + r, where q is an odd integer, and 0 r < n. By (.4) we have M m = µm nq M r ± M nq r. The terms of (M n ) is even, so µm r is an integer. Let d be an integer which divides both M m and M n. Since q is odd, d divides M nq, too. Thus d M nq r holds. On the other hand, if d M n and d M nq r, then similarly d divides M m. Hence gcd(m m, M n ) = gcd(m n, M nq r ). Suppose now m > n and n m. After the first Euclidean-like division by n, replace m by nq r, and continue with this, while the subscript is larger than n. After the last step, nq r might be negative. It is obvious that after two steps m is decreased by 4n. The last term of the sequence coming from these steps depends on the residue of the initial value of m modulo 4n. Let r 1 m (mod n), r m (mod 4n), and 0 < r 1 < n, 0 < r < 4n. In particular, for the last subscript r we found r 1, if 0 < r < n, r n r = 1, if n < r < n, r 1, if n < r < 3n, r 1 n, if 3n < r < 4n. Obviously, gcd(n, r 1 ) = gcd(n, r ) and 0 < r < n, further if d 1 m and d 1 n, then d 1 nq r. Moreover if d 1 divides both n and nq r, then it must divide r and m = nq + r. This shows that gcd(m, n) = gcd(nq r, m). Thus gcd(m, n) = gcd(r, n). Then apply this approach successively (replace the initial values of m by n, and n by r, and continue), and finish when the remainder is zero. The last nonzero remainder is the gcd. To complete the proof of the second case, suppose that gcd(m, n) = 1. By the last division n = 1 follows, and denote the value of m by m 1. The parities of m = nq + r and nq r coincide in each step. If both m and n are odd, then the values of nq r, r are odd, hence so is m 1. If m is even and n is odd, then r is

5 Diophantine triples in a Lucas-Lehmer sequence 9 even, and then the next division-sequence begins with odd m and even n. By the last division (where n = 1) it follows that m 1 must be even. Similarly, if the initial value of m is odd and n is even, then m 1 is even, too. Put d = gcd(m, n). It occurs if we multiply all the terms in the last paragraph by d. If both m/d and n/d are odd, then the quotient in the last division (that is m 1 ) is odd, and by the algorithm and Lemma.1, we have gcd(m m, M n ) = gcd(m m1d, M d ) = M d. If exactly one of m/d and n/d is even, then the last quotient (m 1 ) is even, and gcd(m m, M n ) = gcd(m m1d, M d ) = follows by Lemma.. Now prove the third statement. The explicite formulae provide µl m+n = L n M m + L m M n, (.) µm m+n = 1L n L m + M n M m, (.9) where µ = if both m and n are odd, and µ = 1 otherwise. First we show that gcd(l k, M k ) = if 4 k, and gcd(l k, M k ) = 1 otherwise. It is clear for k = 1,, 3, 4. From (.) and (.9) we obtain By the Euclidean algorithm we have L k+4 = 1 (L km 4 + L 4 M k ) = 7L k + M k, M k+4 = 1 (1L kl 4 + M k M 4 ) = 4L k + 7M k. gcd(l k+4, M k+4 ) = gcd(7l k + M k, 4L k + 7M k ) = gcd(7l k + M k, 3L k + M k ) = gcd(l k, 3L k + M k ) = gcd(l k, M k ). An induction implies the assertion for every k. Now we show gcd(m kn, L n ) = 1 or, again by induction for k. We have just seen that it is true for k = 1. Now (.9) implies µm kn+n = 1L kn L n + M kn M n. Let d be an odd integer such that d M kn+n and d L n. In this case d L kn, and we have shown that gcd(l kn, M kn ), so d is relatively prime to M kn. Thus d M n. Further gcd(l n, M n ), and d is odd, so d = 1. If n is not divisible by 4, then L n is odd, and gcd(m kn+n, L n ) is necessarily 1. If 4 n, then M kn+n is not divisible by 4, but L kn+n is even, so gcd(m kn+n, L n ) =. We will show that if k is odd, then gcd(m n, L kn ) = 1 or. Clearly, it is true for k = 1. Suppose now that it holds for an odd k, and check it for k +. It follows from (.) that µl kn+n = L kn M n + M kn L n. Let be d an odd integer which divides both L kn+n and M n. Then d M kn holds since k is odd. But d is relatively prime to M n, so d must divide L kn. We know

6 90 K. Gueth that gcd(l kn, M kn ), henceforward d = 1. If 4 n, then odd k entails odd L (k+)n, and if 4 n, then 4 M n. Hence gcd(m n, L kn+n ) is 1 or. Assuming k is even, put k = l t, where t is odd. Then M n divides M tn, and we have L tn = µl tn M tn, where µ is 1 or 1/. So M tn / L tn, and by induction, M tn / divides L l tn. Subsequently, gcd(m n, L kn ) is M n or M n / for even k. Thus the third statement is proven if one of n and m divides the other. For general m and n, suppose m > n, and let m = nq + r, where q is odd, 0 < r < n. From (.), µl nq+r = L nq M r + M nq L r follows. It is easy to see that for any odd d the conditions (d L m and d M n ), and (d M n and d M r ) are equivalent (for odd q use that M n divides M nq and gcd(m nq, L nq ) is 1 or ). So it is enough to determine the greatest odd common divisior of M n and M r, for which we use the second part of this lemma. Trivially, gcd(n, r) = gcd(n, m). Denote this value by c. If m/c is even and n/c is odd, then (because q is odd) r/c is odd (say this is case A). By the lemma, gcd(m n, M r ) = M gcd(n,r). If m/c is odd and n/c is even, then r/c is odd. If both m/c and n/c are odd, then r/c is even. In these two cases (we call them case B) gcd(m n, M r ) = hold. Clearly, M n is not divisible by, moreover L m and M n are both divisible by 4 if and only if 4 m and n (mod 4). In this case the exponent of in gcd(n, m) is 1, m/c is even, and n/c is odd (this is case A), and M gcd(n,m) is divisible by 4. It is easy to see that gcd(l m, M n ) = M gcd(n,m). In the remaining situations of case A, M gcd(m,n) is not divisible by 4. Thus gcd(l m, M n ) is M gcd(n,m) or one half of it. In case B, 4 does not divide L m and M n at the same time, so their gcd is 1 or. If m < n, then n = mp + r. Now p is not necessarily odd, therefore we can suppose 0 < r < m. Then from (.9) we conclude gcd(l m, M n ) = gcd(l m, M r ). To complete the proof we must use the previous case of this lemma. The next lemma gives lower and upper bounds on the terms of (L n ) and (M n ) by powers of dominant root α. Lemma.5. Suppose n 3. We have α n < L n < α n 0.943, α n 0.11 < L n+1 < α n 0.10, α n < M n < α n+0.001, α n < M n+1 < α n Further, independently from the parity of the subscript k, hold. α k/ < L k < α k/ 0.60 and α k/ < M k < α k/+0.64 Proof. Let n 0 be a positive integer, and assume n n 0. The explicit formula (.1) simplifies L n = (α n β n )/(α β), which yields L n αn β n0 α β = α n 1 ( β α )n0 α n0 n α β α n 1 ( β α )n0 α β.

7 Diophantine triples in a Lucas-Lehmer sequence 91 Supposing n 0 3, together with 0 < β/α < 1 it leads to 1 ( β α )n0 α β 1 ( β α )3 α β = > α Thus L n > α n To get an upper bound is easier, since β > 0 implies L n = αn β n α β < αn α β = αn 1 3 < αn For odd subscripts a similar treatment is available by First we see L n+1 = 1 [ ( 3 + 1)α n + ( 3 1)β n]. α β Now assume n n 0 3. Consequently, L n+1 > αn > α n L n+1 1 [( 3 + 1)α n + ( ] 3 1)β n0 α β [ ( ) ] n β = α n 3 + α n0 n 3 α α n [ ( β α ) 3 ] = α n < α n The bounds for the terms M n can be shown by an analogous way. Lemma.6. Suppose that a, b, z, and the fractions appearing below are integers. Then 1. if 3a b, then gcd( z+a, 3z+b ) 3a b,. if a b, then gcd( z+a, z+b 6 ) a b, 3. if a b, then gcd( z+a, z+b 4 ) a b. Proof. The statements follow by a simple use of the Euclidean algorithm. Lemma.7. Supposing z 4, the following properties are valid. 1. If z 1 (mod 4), then M z 1. If z 3 (mod 4), then M z 1 3. If z (mod 4), then M z < L z, further 3L z 1 < 4L z. < L z. < L z.

8 9 K. Gueth 4. If z 0 (mod 4), then M z < 4L z. Proof. Use (.5), (.6), and M n = { L n 1 + L n+1, if n is even, (L n 1 + L n+1 ), if n is odd. (.10) Here (.10) can be proven by induction. Lemma.. Suppose that a and b are positive real numbers and u 0 is a positive integer. Let κ = log α (a + b α ). If u u u 0 0, then aα u + b α u+κ. Proof. This is obvious by an easy calculation. 3. Proof of Theorem 1.1 The conditions 1 a < b < c entail 3 x < y < z. Obviously, c L y 1 and c L z 1. Thus c gcd(l y 1, L z 1). Clearly, L z = bc + 1 < c, which implies Lz < c. Combining this with Lemma.5, we see α z = α 1 ( z 0.944) < L z < c < L y < α y 0.60, and then z/ < y/ 0.60 yields z < y 0.3. Hence z y 1. Now we distinguish two cases. Case I: z 117. The key point of this case is to estimate G = gcd(l y 1, L z 1). Assume that i, j {±1, ±}, and µ i, µ j {1, 1/}. By Lemma.3, G = gcd(µ i L y i gcd(l y i gcd(l y i M y+i M y+i, L z j, µ j L z j, L z j M z+j ) gcd(l y i M z+j ) ), M z+j Let Q denote the last product. By Lemma.4 ) gcd(m y+i, L z j ) gcd(m y+i Q L gcd( y i, z j )M gcd( y i, z+j )M gcd( y+i, z j )M gcd( y+i, z+j ) follows. We define d 1, d, d 3, d 4 according to the relations ( y i gcd, z j ) = z j ( y i, gcd d 1, z + j ) ( y + i gcd, z j ) = z j ( y + i, gcd d 3, z + j ) = z + j d, = z + j d 4., M z+j ).

9 Diophantine triples in a Lucas-Lehmer sequence 93 Let d = min{d 1, d, d 3, d 4 }. First suppose d 5. Now Lemma.5, together with i, j implies α z < Q L z j d M z+j d M z j M z+j d d L z j 10 M z+j 10 < α z ( α z ) 3 = α z M z j M z+j But z/ < (z + )/ contradicting z 117. Now let d = 4, that is one of d 1, d, d 3, d 4 equals 4. Assume that η 1, η {±1}. Then η 1 j, η i, and we can assume z + η 1 j y + η i. Contrary, if it does not hold, then by the definition of d the inequality 5/4(z ) y + is true, which together with z > y implies 5z 4y + 1 < 5y + 1. So z < 1, which is not the case. Now we have only two possibilities: z + η 1 j = y + η i or z + η 1 j = y + η i. 6 In the first case we have z = 4y + (4η i η 1 j) 4y 10, and by z y 1 we get 4y 10 y 1, which implies y 4, and then z 7, a contradiction. In the second case let η 1, η {±1}, such that (η 1, η ) (η 1, η ). Clearly, y = 3z + 3η 1j 4η i, and 4 y + η i = 3z + 3η 1j + 4(η η )i. Put t = 4(η η ). Thus t = 0 or ±. Applying the first assertion of Lemma.6 with a = η 1j and b = 3η 1 j + ti, it gives ( ) ( ) z + η gcd 1j, y + η i z + η = gcd 1j, 3z + 3η 1j + ti 3η 1j 3η 1 j ti, which does not exceed 14. This conclusion is correct if 3a b 0, that is if 3η 1 3η 1 j ti 0. If 3a b = 0, then 3 t, and then t = 0. Thus η 1 must be equal to η 1, so (η 1, η ) = (η 1, η ), which has been excluded. Subsequently, three of the four factors of Q is at most M 14 (M n L n for any index n) and the fourth factor is L z±j or M z±j, none of them exceeding M z+. So Q M 3 14M z+ and then, by Lemma.5, we have = M z+, α z < Q < α α z Now we conclude z < 116.7, and it is a contradiction with z 117. Suppose d = 3. We have the two possibilities z + η 1 j 6 = y + η i and z + η 1 j 6 = y + η i. 4

10 94 K. Gueth In the first case y 1 z = 3(y + η i) η 1 j 3y implies y 7, and then z 13, which is impossible. In the second case we repeat the treatment of case d = 4, the variables η 1 and η satisfy the same conditions. Now y = (z + η 1 j 3η i)/3 provides y + η i = z + η 1j 3η i + 3η i 6 = z + η 1j + 3(η η )i. 6 Let be t = 3(η η ) with value 0 or ±6. Use the second assertion of Lemma.6 with a = η 1j, b = η 1 j + ti. If a b 0 then ( ) ( ) z + η gcd 1j, y + η i z + η = gcd 1j, z + η 1j + ti η 1j η 1 j ti 6, which is less then or equal to 10. If a b = 0, that is if η 1j η 1 j ti = 0, then 3 t and j t show 3 η 1 η 1, which can hold only if η 1 = η 1. But in this case t must be zero, too. So (η 1, η ) = (η 1, η ), which is not allowed. We have α z < Q M 3 10M z+ 6 < 74 3 α z by using Lemma.5. This implies z < 96, again a contradiction. Now suppose d =. The only possibility is z + η 1 j 4 = y + η i. (η 1 and η are the same as in the previous cases.) It leads to y = (z +η 1 j η i)/, and then to y + η i = z + η 1j η i + η i 4 = z + η 1j + ti, 4 where t = (η η ) {0, ±4}. Let a = η 1j, b = η 1 j + ti. If a b, then by the third assertion of Lemma.6 we have ( ) ( ) z + η gcd 1j, y + η i z + η = gcd 1j, z + η 1j + ti η 1j η 1 j ti 4 6. Thus α z < Q M 3 6 M z+ 4 < α α z , and we arrived at a contradiction via z < 0. If a b = 0, then (η 1 η 1 )j = ti. Now, if j = ±1, then (because t is divisible by 4) 4 η 1 η 1 must hold. This occurs only if η 1 = η 1, hence t = 0, so η = η, which has been excluded. Thus we may suppose j = ± and η 1 η 1. In this case η 1 η 1 = ±, and i = ±1. The factors of Q belong to ( η 1, η ) and (η 1, η ) can be estimated by M 6. If (η 1, η ) = (1, 1), then this factor is gcd(m y+i, M z+j ), which is via (z + j)/4 = (y + i)/ and

11 Diophantine triples in a Lucas-Lehmer sequence 95 Lemma.4. If (η 1, η ) = (1, 1), then similarly gcd(l y i, M z+j ). In this two cases we have α z < Q M 6 M z+ 4 < α 6.57 α z , and then z 60, a contradiction. Let (η 1, η ) = ( 1, 1) or ( 1, 1). From (z + η 1 j)/4 = (y + η i)/ and j =, i = 1 it is easy to see that (z η 1 j)/ = (y η i)/ or (z η 1 j)/ = (y η i)/ ± 4. If the first case holds, then gcd((z η 1 j)/, (y η i)/) = (z η 1 j)/4. Further if (η 1, η ) = ( 1, 1), then the factor of Q belonging to ( η 1, η ) is gcd(m y+i, M z+j ) = (by Lemma.4). If (η 1, η ) = ( 1, 1), then the factor gcd(l y i, M z+j ) = 1 or. If (z η 1 j)/ = (y η i)/ ± 4 holds, it can be seen by the Euclidean algorithm that gcd((z η 1 j)/, (y η i)/) 4, and the factor of Q is at most M 4 = 14. So in these cases we conclude and this implies z < 7. Assume d = 1. Now α z < Q M 4 M 6 M z+ 4 z + η 1 j = y + η i, < α.005 α z , where η 1, η = ±1, and it reduces to z ± j = y ± i with i, j {±1, ±} According to Lemma.3 the values depend of the residue y and z modulo 4. Altogether, it means that we need to verify 16 cases. 1. y z 1 (mod 4). Clearly, now i = j = 1, so z ± 1 = y ± 1. The condition y z (mod 4) leads immediately to y = z, a contradiction.. y 1, z (mod 4). Now i = 1, j =. Thus z ± = y ± 1, and then z = y ±3 or z = y ±1. Considering them modulo 4, the only possibility is z = y +1. By Lemma.3, we conclude L y 1 = L y 1 M y+1 = L z M z, and L z 1 = L z M z+. The common factor L z together with gcd(m z, M z+ ) = and by Lemma.5 provides a contradiction again, since α z < gcd(l y 1, L z 1) = L z < α 0.57 α z = α z y 1, z 3 (mod 4). Here i = 1, j = 1, and the only possibility is z = y +. It follows that where gcd(l z+1 L y 1 = L y 1 M y+1, L z 3 ) = 1. Now = L z 3 M z 1, L z 1 = L z+1 M z 1, c gcd(l y 1, L z 1) = M z 1 = c 1 c > c 1 Lz

12 96 K. Gueth holds with an appropriate integer c 1. By Lemma.7, M z 1 c 1 Lz < M z 1 < L z. So we have < L z, which implies c 1 <, i.e. c 1 = 1. Thus c = M z 1 we can see from the factorization of L y 1 and L z 1 that a = L z 3 Lemma.5 shows α x 0.60 > L x = ab + 1 = L z 3 L z+1, and., b = L z > L z 3 L z+1 > α z α z Clearly, x > z 3.416, and then x z 3. In our case x < y = z holds, so x = z 3. This implies L z 3 1 = L x 1 = L z 3 L z+1, which entails L z 3 L z 3 1. Combining it with L z 3 L z 3, we have L z 3 = 1, and z is too small. 4. y 1, z 0 (mod 4). In this case z = y + 3, and L y 1 = L y 1 M y+1 = L z 4 M z, L z 1 = 1 L z+ M z. The distance of the subscripts of the appropriate terms of (L n ) is 3, so gcd(l z 4, 1 L z+ ) gcd(l z 4, L z+ ) = 1 or 3. So gcd(l y 1, L z 1) 3M z. Therefore there exist a positive integer c 1 such that c gcd(l y 1, L z 1) 3M z = c 1 c > c 1 Lz. Lemma.7 implies M z c 1 < 6. Since L z+ 1, L z 1) = λm z < L z, and so 6 L z > 3M z > c 1 Lz hold. Thus is odd, M z does not divide L z 1. So we have gcd(l y /, where λ = 1 or 3. / = 3M z /6, which implies c 1 6, a contradic- tion. When λ = 1, c divides M z Assuming λ = 3, it yields c 3M z c = 3M z /. Thus either c = 3M z / (c 1 = ) or /4 (c 1 = 4) holds. We can exclude the second case, because (z )/ is not divisible by 4. In the first case b = L z+ /3 and /3 follow from is odd, and so M z a = L z 4 bc = L z 1 = 1 M z L z+ and ac = L y 1 = M z L z 4, respectively. Using the fact that L k L k = L k 1 L k holds for every positive integer k (this comes from the explicit formula (.1)), we can write L x = ab + 1 = 9 L z 4 L z+ + 1 = (L z L z 9 By Lemma.5 we obtain α x 0.60 > L x = 9 L z L z > 9 L z 1) + 1 = 9 L z L z L z > α α z α z (since (z )/ is odd). It implies x > z 5.176, so x z 5 holds.

13 Diophantine triples in a Lucas-Lehmer sequence 97 We will reach the contradiction by showing ab + 1 < L z 5. Knowing that z is even, L z 5 > α z = α z 3.11 follows from Lemma.5. Since L z L z > α z α z = α z.15 and z 16, the exponent of α is at least Applying Lemma. with u 0 = 5, we have κ = log α (( + 7α 5 )/9) < 1.13, and then ab + 1 = 9 L z L z < α 1.13 α z α z = α z From these inequalities L z 5 > α z 3.11 > α z 3.61 > ab + 1 follows, and the proof of this part is complete. 5. y, z 1 (mod 4). Now z = y + 3, further L y 1 = L y M y+ = L z 5 M z 1 It is easy to see from Lemma.4 that gcd(l z 5 gcd(l z 5, M z+1 ) M 3 = 10, gcd(m z 1, L z 1, L z 1 = L z 1, L z 1 ). Consequently, M z+1. ) = 1, gcd(m z+1 α z < gcd(l y 1, L z 1) 40 < α.0, and then z < 14, a contradiction again., M z 1 ) =, 6. y z (mod 4). In this case i = j =. Then z = y + 4 follows. The identities and gcd(l z 6 L y 1 = L y M y+, L z divisible by 4), gcd(l z 6.4) induce which gives z < 15. = L z 6 ) = 1, gcd(m z, M z+ M z, M z+ ) M 4 = 14, gcd(m z, L z 1 = L z M z+ ) = (because both terms cannot be, L z ) (see Lemma α z < gcd(l y 1, L z 1) 56 < α 3.057, 7. y, z 3 (mod 4). Here z = y + 1, moreover we have Again by Lemma.4, L y 1 = L y M y+ gcd(l z 3 gcd(l z 3, L z+1, M z 1 = L z 3 M z+1 ) = 1, gcd(m z+1 ), gcd(m z+1, L z 1 = L z+1, M z 1 ) =, )., L z+1 M z 1.

14 9 K. Gueth Thus follows, which implies z < 9. α z < gcd(l y 1, L z 1) < α y, z 0 (mod 4). Now i =, j =, and y ± = j cannot hold modulo y 3, z 1 (mod 4). In this case the only possibility is z = y +. Obviously, L y 1 = L y+1 M y 1 = L z 1 M z 3 hold. Beside the common factor, we get gcd(m z 3 ) = (because the subscripts are odd). Hence gcd(l y 1, L z 1) = L z 1, L z 1 = L z 1 M z+1, M z+1, further we see c gcd(l y 1, L z 1) = L z 1 = c 1 c > c 1 Lz with an appropriate c 1. By the second assertion of case (1) in Lemma.7, L z > 3/L z 1, subsequently L z 1 > c 1 Lz > c 1 3 L z 1 holds, providing c 1 < 3 from the factorizations <. So only c 1 = 1 is possible. Thus c = L z 1, and we obtain ac = L y 1 = L z 1 a = 1 M z 3 M z 3, bc = L z 1 = L z 1 M z+1 and b = 1 M z+1. Finally, we show that c < b. (.10) yields M k+1 = L k + L k+ > 4L k. Now (z 1)/ is even, so L z 1 < 1 M z+1. Thus c < b, contradicting the condition a < b < c. 10. y 3, z (mod 4). We find z = y + 3, and L y 1 = L y+1 M y 1 By Lemma.4, gcd(m z 4 = L z M z 4, L z 1 = L z M z+., M z+ ) = follows (not M 3 = 10, because if the sub- scripts are divisible by 3, dividing them by 3 exactly one of the integers will be odd). Now α z < gcd(l y 1, L z 1) = L z < α 0.57 α z leads to a contradiction.

15 Diophantine triples in a Lucas-Lehmer sequence y z 3 (mod 4). In this case, i = j = 1 implies y = z, which is a contradiction. 1. y 3, z 0 (mod 4). Here z = y + 1, further L y 1 = L y+1 M y 1 = L z M z, L z 1 = 1 L z+ M z hold. Lemma.4 provides gcd(l z, L z+ ) = 1, and we obtain gcd(l y 1, L z 1) = 1 M z (because L z+ is odd). Hence c gcd(l y 1, L z 1) = 1 M z = c 1 c > c 1 Lz. By Lemma.7 we have M z < L z. Thus M z implies c 1 < 1, an impossibility. 13. y 0, z 1 (mod 4). In this case z = y + 1, moreover L y 1 = 1 L y+ M y By Lemma.4, we obtain gcd(l z+1 gcd(l z+1, M z+1 ), gcd(m z 3 implies z < 9. = 1 L z+1, L z 1, L z 1 M z 3 > c 1 Lz > c 1 M z, which, L z 1 = L z 1 ) = 1, gcd(m z 3 ). Then α z < gcd(l y 1, L z 1) < α M z+1., M z+1 ) =, 14. y 0, z (mod 4). Now, by Lemma.3, i =, j =, and y = z ± follow, which is not possible. 15. y 0, z 3 (mod 4). In this case z = y + 3, and L y 1 = 1 L y+ M y Via Lemma.4 we see gcd(l z 1 = 1 L z 1, L z+1 gcd(l z 1, M z 1 ) = 1, (because z 1 M 3 = 10. These lead to a contradiction via M z 5 ) = 1, gcd(m z 5, and so L z 1, L z 1 = L z+1, M z 1 ) =, M z 1. is odd), gcd(m z 5 α z < gcd(l y 1, L z 1) 0 < α.75., L z+1 ) 16. y z 0 (mod 4). In the last case the only possibility is z = y + 4. We have L y 1 = 1 L y+ M y By Lemma.4, we get gcd(l z = 1 L z, L z+ ) = 1, M z 6, L z 1 = 1 L z+ M z.

16 100 K. Gueth gcd(m z 6, M z ) =, gcd(l z gcd(m z 6 Then we obtain z < 10 from, M z, L z+ ) = 1 (because (z )/ is odd), ) M 4 = 14. α z < gcd(l y 1, L z 1) 14 < α.004. Case II: z 116. The proof of Theorem 1 will be complete, if we check the finitely many cases 3 x < y < z 116. It has been done by a computer verification based on the following observation. The equations (1.) imply Thus must be an integer. integer. References (L x 1)(L y 1) = a bc = a (L z 1). (L x 1)(L y 1) L z 1 (3.1) Checking the given range we found that (3.1) is never an [1] M. Alp, N. Irmak and L. Szalay, Balancing Diophantine Triples, Acta Univ. Sapientiae 4 (01), [] M. Alp and N. Irmak, Pellans sequence and its diophantine triples, Publ. Inst. Math. 100(114) (016), [3] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 (004), [4] C. Fuchs, F. Luca and L. Szalay, Diophantine triples with values in binary recurrences, Ann. Scuola Norm. Sup. Pisa. Cl. Sci. III 5 (00), [5] C. Fuchs, C. Huttle, N. Irmak, F. Luca and L. Szalay, Only finitely many Tribonacci Diophantine triples exist, accepted in Math. Slovaca. [6] C. Fuchs, C. Huttle, F. Luca, L. Szalay, Diophantine triples and k-generalized Fibonacci sequences, accepted in Bull. Malay. Math. Sci. Soc. [7] B. He, A. Togbe and V. Ziegler, pdf. [] V. E. Hoggatt and G. E. Bergum, A problem of a Fermat and Fibonacci sequence, Fibonacci Quart. 15 (1977), [9] F. Luca and L. Szalay, Fibonacci Diophantine Triples, Glasnik Math. 43(63) (00), [10] F. Luca and L. Szalay, Lucas Diophantine Triples, Integers 9 (009),