Diophantine triples in a Lucas-Lehmer sequence

Annales Mathematicae et Informaticae 49 (01) pp. 5 100 doi: 10.33039/ami.01.0.001 http://ami.uni-eszterhazy.hu Diophantine triples in a Lucas-Lehmer sequence Krisztián Gueth Lorand Eötvös University Savaria Department of Mathematics Károli Gáspár tér 4 9700 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

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 = 1 + 4 3 (ωn ψ n ) + 1 4 3 (( 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

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)

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) + 1 1 1 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

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

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 0.944 < 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+0.763 < M n+1 < α n+0.764. Further, independently from the parity of the subscript k, hold. α k/ 0.944 < 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.

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 = 0.56... > α 0.944. Thus L n > α n 0.944. 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 0.943. 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 > 1 + 3 3 αn > α n 0.11. L n+1 1 [( 3 + 1)α n + ( ] 3 1)β n0 α β [ ( ) ] n0 3 + 1 3 1 β = α n 3 + α n0 n 3 α α n [ 3 + 1 3 + 3 1 3 ( β α ) 3 ] = α n 0.7753... < α n 0.10. 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.

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 4 0.47 = α 1 ( z 0.944) < L z < c < L y < α y 0.60, and then z/4 0.47 < 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 ).

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 4 0.47 < 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+ 0 0.60 ( α z+ 0 +0.64) 3 = α z+ 5 +0.11. M z j M z+j 10 10 But z/4 0.47 < (z + )/5 + 0.11 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 = 1004 3 M z+, α z 4 0.47 < Q < α 1.003 α z+ 16 +0.64. 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

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 4 0.47 < Q M 3 10M z+ 6 < 74 3 α z+ 1 +0.64 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 4 0.47 < Q M 3 6 M z+ 4 < α 9.003 α z+ +0.64, 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

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 4 0.47 < Q M 6 M z+ 4 < α 6.57 α z+ +0.64, 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 4 0.47 < Q M 4 M 6 M z+ 4 z + η 1 j = y + η i, < α.005 α z+ +0.64, 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 4 0.47 < gcd(l y 1, L z 1) = L z < α 0.57 α z 4 0.60 = α z 4 0.653. 3. 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

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+1 + 1 > L z 3 L z+1 > α z 3 4 0.944 α z+1 4 0.944. 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+1 + 1 = 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 + 7 9 > 9 L z 1) + 1 = 9 L z L z + 7 9. L z > α 1.143 α z 4 0.61 α z 4 0.944 (since (z )/ is odd). It implies x > z 5.176, so x z 5 holds.

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 5 0.61 = α z 3.11 follows from Lemma.5. Since L z L z > α z 4 0.61 α z 4 0.944 = α z.15 and z 16, the exponent of α is at least 5.75. 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 + 7 9 < α 1.13 α z 4 0.6 α z 4 0.943 = α z 3.61. 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 4 0.47 < 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 4 0.47 < 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.

9 K. Gueth Thus follows, which implies z < 9. α z 4 0.47 < gcd(l y 1, L z 1) < α 1.579. y, z 0 (mod 4). Now i =, j =, and y ± = j cannot hold modulo 4. 9. 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 4 0.47 < gcd(l y 1, L z 1) = L z < α 0.57 α z 4 0.60 leads to a contradiction.

Diophantine triples in a Lucas-Lehmer sequence 99 11. 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 4 0.47 < gcd(l y 1, L z 1) < α 1.579 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 4 0.47 < 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.

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 4 0.47 < 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), 11 19. [] M. Alp and N. Irmak, Pellans sequence and its diophantine triples, Publ. Inst. Math. 100(114) (016), 59-69. https://doi.org/10.9/pim161459i [3] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 (004), 13 14. https://doi.org/10.1515/crll.004.003 [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), 579 60. [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, http://vziegler.sbg.ac.at/papers/quintuple. pdf. [] V. E. Hoggatt and G. E. Bergum, A problem of a Fermat and Fibonacci sequence, Fibonacci Quart. 15 (1977), 33 330. [9] F. Luca and L. Szalay, Fibonacci Diophantine Triples, Glasnik Math. 43(63) (00), 53 64. https://doi.org/10.3336/gm.43..03 [10] F. Luca and L. Szalay, Lucas Diophantine Triples, Integers 9 (009), 441 457. https://doi.org/10.1515/integ.009.037