Finding all S -Diophantine quadruples for a fixed set of primes S
aa r X i v : . [ m a t h . N T ] O c t FINDING ALL S -DIOPHANTINE QUADRUPLES FOR A FIXEDSET OF PRIMES S VOLKER ZIEGLER
Abstract.
Given a finite set of primes S and a m -tuple ( a , . . . , a m ) of pos-itive, distinct integers we call the m -tuple S -Diophantine, if for each 1 ≤ i Given an irreducible polynomial f ∈ Z [ X, Y ] and a set A of positive integers weconsider the product Π = Y a,b ∈ Aa = b f ( a, b ) . It is an interesting question what is the largest prime divisor P (Π) of Π or alterna-tively what is the number of prime divisors ω (Π) of Π in terms of the size of | A | . Incase of f ( x, y ) = x + y this question was intensively studied by several authors start-ing with Erd˝os and Tur´an [4] who established that in this case ω (Π) ≫ log | A | . Itwas shown by Erd˝os, Stewart and Tijdeman [3] that this is essentially best possible.In the case of f ( x, y ) = xy + 1 the investigation was started by S´ark¨ozy [8]. Itwas shown by Gy˝ory et.al. that ω (Π) ≫ log | A | . However it seems that this lowerbound is far from being best possible and the situation is not so clear as in the case f ( x, y ) = x + y .One recent approach to this problem was picked up by Szalay and Ziegler [11]who studied so-called S -Diophantine m -tuples. Let S be a finite set of primes and( a , . . . , a m ) a m -tuple of positive, distinct integers with a < · · · < a m . Then wecall ( a , . . . , a m ) S -Diophantine, if for each 1 ≤ i < j ≤ m the quantity a i a j + 1has prime divisors coming only from the set S . In other words if A = { a , . . . , a m } and Π has only prime divisors coming form S , then ( a , . . . , a m ) is called a S -Diophantine m -tuple. Szalay and Ziegler [11] conjectured that if | S | = 2, then no S -Diophantine quadruple exists. This conjecture has been confirmed for severalspecial cases by Luca, Szalay and Ziegler [11, 12, 13, 16, 5]. Even a rather efficientalgorithm has been described by Szalay and Ziegler [13] that finds for a given set S = { p, q } of two primes all S -Diophantine quadruples, if there exist any. However,the algorithm is strictly limited to the case that | S | = 2 and does not work (norany slightly modified version of the algorithm) for | S | ≥ Mathematics Subject Classification. Key words and phrases. Baker’s method, S -unit equations, S -Diophantine tuples.The author was supported by the Austrian Science Fund(FWF) under the project I4406. In the present paper we establish an algorithm that finds for any set S = { p, q, r } of three primes all S -Diophantine quadruples. Unfortunately the algorithm is notso efficient as its counter part for | S | = 2. However, when implemented in Sage [14]it takes on a usual PC (Intel i7-8700) – using a single core – about six and a halfhours to verify: Theorem 1. There is no { , , } -Diophantine quadruple. However the case S = { , , } is in some sense the slowest instance for smallprimes p, q and r . Thus within four days of runtime we could establish the followingresult: Theorem 2. Let S = { p, q, r } with ≤ p < q < r ≤ . Then all S -Diophantinequadruples are listed in Table 1. S Quadruples { , , } (1 , , , { , , } (1 , , , { , , } (1 , , , Table 1. S -Diophantine quadruples.Thus Theorem 2 raises the following problem: Problem 1. Are the S -Diophantine quadruples listed in Table 1 all S -Diophantinequadruples with | S | = 3 ? Although we present the method only in the case that | S | = 3 obvious modifi-cations to the algorithm would provide an algorithm that would work for any set S of primes with | S | ≥ 3. However, already in the case that | S | = 4 the algorithmwould take instead of several hours several months of computation time. Thus asystematic search for S -Diophantine quadruples in the case | S | = 4 seems to be notfeasible. Therefore we refer from discussing the case | S | > S -Diophantine quadruples. In Section 3 we establish several results that willallow us to find small upper bounds for the relevant unknowns and will also allowus to apply LLL-reduction to reduce these bounds. The LLL-reduction will bediscussed in Section 4. In Section 5 we will give details how to apply the LLL-reduction method to find upper bounds small enough to enumerate all possible S -Diophantine quadruples for a given set S = { p, q, r } of three primes.2. Auxiliary results Assume that ( a, b, c, d ) is a S -Diophantine quadruple, with S = { p, q, r } . Inparticular, we assume that a < b < c < d . Moreover, we will assume that p < q < r .Let us write ab + 1 = s = p α q β r γ bc + 1 = s = p α q β r γ ac + 1 = s = p α q β r γ bd + 1 = s = p α q β r γ ad + 1 = s = p α q β r γ cd + 1 = s = p α q β r γ . -DIOPHANTINE QUADRUPLES 3 The almost trivial observation that ab · cd = ( s − s − 1) = ( s − s − 1) = ac · bd yields the (non-linear) S -unit equation s s − s − s = s s − s − s . Similar computations lead us to the following system of S -unit equations s s − s − s = s s − s − s ,s s − s − s = s s − s − s ,s s − s − s = s s − s − s . (1)Let us call two indices i, j ∈ { , . . . , } complementary, if i + j = 7. Note thatif i and j are complementary, then we have ( s i − s j − 1) = abcd . With thesenotations at hand the following lemmas can be proved (e.g. see [11]) by elementarymeans. Lemma 1. The smallest two members of each of the sets { α , α , α , α } , { α , α , α , α } , { α , α , α , α } are equal. Similar statements also hold for the β ’s and γ ’sProof. This follows by p -adic considerations of the system of equations (1) (cf. [11,Proposition 1]). (cid:3) Lemma 2. Assume that ( a, b, c, d ) is a S -Diophantine quadruple. Then we have a | gcd (cid:18) s − s gcd( s , s ) , s − s gcd( s , s ) , s − s gcd( s , s ) (cid:19) ,b | gcd (cid:18) s − s gcd( s , s ) , s − s gcd( s , s ) , s − s gcd( s , s ) (cid:19) ,c | gcd (cid:18) s − s gcd( s , s ) , s − s gcd( s , s ) , s − s gcd( s , s ) (cid:19) ,d | gcd (cid:18) s − s gcd( s , s ) , s − s gcd( s , s ) , s − s gcd( s , s ) (cid:19) . Proof. A proof can be found in [11, Lemma 3]. (cid:3) Lemma 3. Let ( a, b, c ) be a S -Diophantine triple with ab + 1 = s = p α q β r γ ac + 1 = s = p α q β r γ bc + 1 = s = p α q β r γ . Then we have: • If p = 2 , then min { α , α , α } ≤ . • If p ≡ , then min { α , α , α } = 0 . • If q ≡ , then min { β , β , β } = 0 . • If r ≡ , then min { γ , γ , γ } = 0 . VOLKER ZIEGLER Proof. A proof can be found in [12, Lemma 2.2] except for the case that p = 2.However, in the case that p = 2 we slightly change the proof. That is we assumethat min { α , α , α } ≥ a b c = ( s − s − s − ≡ − ≡ (cid:3) In order to obtain a first upper bound in the next section we will apply resultson lower bounds for linear forms in (complex and p -adic) logarithms. Therefore let η = 0 be an algebraic number of degree δ and let a ( X − η ) · · · ( X − η δ ) ∈ Z [ X ]be the minimal polynomial of η . Then the absolute logarithmic Weil height isdefined by h ( η ) = 1 δ log | a | + δ X i =1 max { , log | η i |} ! . We will mainly need the notation of height for rational numbers. Therefore let usremark that if p/q ∈ Q with p, q integers such that gcd( p, q ) = 1, then h ( p/q ) =max { log | p | , log | q |} . With this basic notation we have the following result on lowerbounds for linear forms in (complex) logarithms due to Matveev [6]. Lemma 4. Denote by η , . . . , η n algebraic numbers, nor neither , by log η , . . . , log η n determinations of their logarithms, by D the degree over Q of the numberfield K = Q ( η , . . . , η n ) , and by b , . . . , b n rational integers. Furthermore let κ = 1 if K is real and κ = 2 otherwise. Choose A i ≥ max { Dh ( η i ) , | log α i |} (1 ≤ i ≤ n ) and E = max { , max {| b j | A j /A n : 1 ≤ j ≤ n }} . Assume that b n = 0 and log η , . . . , log η n are linearly independent over Z . Then log | b log η + · · · + b n log η n | ≥ − C ( n ) C W D Ω , with Ω = A · · · A n ,C ( n ) = C ( n, κ ) = 16 n ! κ e n (2 n + 1 + 2 κ )( n + 2)(4( n + 1)) n +1 (cid:18) en (cid:19) κ ,C = log (cid:0) e . n +7 n . D log( eD ) (cid:1) , W = log(1 . eED log( eD )) . In our case of main interest where | S | = 3 we can apply results on linear forms intwo p -adic logarithms. For a prime p we denote by Q p the field of p -adic numberswith the standard p -adic valuation ν p . As above let η , η be algebraic numbersover Q and we regard them as elements of the field K p = Q p ( η , η ). In our case η and η will be rational integers, thus K p = Q p . Similar as in the case of Matveev’stheorem above we have to use a modified height. In particular, we write h ′ ( η i ) ≥ max { h ( η i ) , log p } , ( i = 1 , . With these notations at hand, let us state the result due to Bugeaud and Lau-rent [1, Corollary 1]): -DIOPHANTINE QUADRUPLES 5 Lemma 5. Let b , b be positive integers and suppose that η and η are multiplica-tively independent algebraic numbers such that ν p ( η ) = ν p ( η ) = 0 . Put E ′ := b h ′ ( η ) + b h ′ ( η ) . and E := max { log E ′ + log log p + 0 . , , 10 log p } . Then we have ν p ( η b η b − ≤ p (log p ) E h ′ ( η ) h ′ ( η ) . The next lemma is an elementary, but rather useful result due to Peth˝o and deWeger [7]. For a proof of Lemma 6 we refer to [9, Appendix B]. Lemma 6. Let u, v ≥ , h ≥ and x ∈ R be the largest solution of x = u + v (log x ) h .Then x < max { h ( u /h + v /h log( h h v )) h , h ( u /h + 2 e ) h } . A first upper bound As already mentioned in the previous section we consider the case that | S | = 3,i.e. we have S = { p, q, r } with P = max { p, q, r } and write ab + 1 = s = p α q β r γ bc + 1 = s = p α q β r γ ac + 1 = s = p α q β r γ bd + 1 = s = p α q β r γ ad + 1 = s = p α q β r γ cd + 1 = s = p α q β r γ . Let us write A = max i =1 ,..., { α i } , B = max i =1 ,..., { β i } and C = max i =1 ,..., { γ i } and M = max { A, B, C } .The main result of this section is the following proposition: Proposition 1. If there exists an S -Diophantine quadruple, then M < . · (log P ) log(3 . · (log P ) ) . In the case that S = { , , } we have the upper bound M < . · and in the case that S = { p, q, r } with p, q, r < we have the upper bound M < . · . First, we prove an upper bound for α i , β i and γ i with i = 1 , , Lemma 7. We have log s , log s ≤ . · log p log q log r log(2 M ) and log s ≤ . · log p log q log r log(2 M ) . In particular, we have α , α ≤ . · log q log r log(2 M ) ,β , β ≤ . · log p log r log(2 M ) ,γ , γ ≤ . · log p log q log(2 M ) , VOLKER ZIEGLER and α ≤ . · log q log r log(2 M ) ,β ≤ . · log p log r log(2 M ) ,γ ≤ . · log p log q log(2 M ) . Proof. Following Stewart and Tijdeman [10] we consider the quantity(2) T = p A ′ q B ′ r C ′ = ( ab + 1)( cd + 1)( ac + 1)( bd + 1) = 1 + cd + ab − ac − bd ( ac + 1)( bd + 1) = 1 + θab + 1with | θ | < A ′ = α + α − α − α , B ′ = β + β − β − β and C ′ = γ + γ − γ − γ .Note that we have | A ′ | ≤ A , | B ′ | ≤ B and | C ′ | ≤ C . Let us note that log(1+ x ) ≤ x provided that | x | ≤ / 2. Since ab + 1 ≥ | Λ | = | A ′ log p + B ′ log q + C ′ log r | < p α q β r γ = 2 s We apply Matveev’s theorem (Lemma 4) with D = 1 , n = 3 , κ = 1 , b = A ′ , b = B ′ , b = C ′ ,η = p, η = q, η = r, A = log p, A = log q, A = log r. Therefore we have E ≤ max { A ′ , B ′ , C ′ } ≤ M provided M > s − log 2 < . · log p log q log r log(2 M ) , which yields the upper bound for log s . Sincelog s = α log p + β log q + γ log r this also yields the upper bounds for α , β and γ .If we consider instead of T the quantity(4) T = p A ′′ q B ′′ r C ′′ = ( ac + 1)( bd + 1)( bc + 1)( ad + 1) = 1 + bd + ac − bc − ad ( bc + 1)( ad + 1) = 1 + θac + 1with | θ | < A ′′ = α + α − α − α , B ′′ = β + β − β − β and C ′′ = γ + γ − γ − γ .We end up with the linear form(5) | Λ | = | A ′′ log p + B ′′ log q + C ′′ log r | < p α q β r γ and again an application of Matveev’s result yields the same upper bound for log s and also the same upper bounds for α , β and γ .Finally let us note that s = bc + 1 < ( ab + 1)( ac + 1) = s s , which yields after some easy computations the stated upper bounds for log s , α , β and γ . (cid:3) Let us denote by M and M upper bounds for max { log s , log s } and log s respectively. Then the previous lemma provides upper bounds for M and M . Lemma 8. We have M ≤ . · M (log P ) log M . -DIOPHANTINE QUADRUPLES 7 Proof. Following again the arguments of Stewart and Tijdeman [10] we considerthe following inequality0 ≤ ba p α − α q β − β r γ − γ − ba · ad + 1 bd + 1 − b − aa ( bd + 1) ≤ d which implies(6) Λ := (cid:12)(cid:12)(cid:12)(cid:12) log ba + A ′ log p + B ′ log q + C ′ log r (cid:12)(cid:12)(cid:12)(cid:12) ≤ d , where A ′ = α − α , B ′ = β − β and C ′ = γ − γ . We apply Matveev’s resultwith D = 1 , n = 4 , κ = 1 , b = 1 , b = A ′ , b = B ′ , b = C ′ ,η = ab , η = p, η = q, η = r,A = log (max { a, b } ) , A = log p, A = log q, A = log r. Since b < ab + 1 = exp( α log p + β log q + γ log r ) ≤ exp( M )and max { A ′ , B ′ , C ′ } ≤ M we get1 . · M log p log q log r log M > log d − log 2 . On the other hand we have d > dc + 1 ≥ exp (max { A log p, B log q, C log r } ) ≥ exp( M min { log p, log q, log r } ) . Combining these two inequalities yields the content of the lemma. (cid:3) Now the proof of Proposition 1 is a combination of Lemmas 7 and 8. Proof of Proposition 1. By inserting the bound for M obtained by Lemma 7 intothe bound provided by Lemma 8 we obtain the following inequality M < . · (log P ) (log 2 M ) . An application of Lemma 6 yields M < . · (log P ) log(3 . · (log P ) ) . If we put P = 5 and P = 100 we obtain the other bounds stated in the proposi-tion. (cid:3) Although we have an upper bound for M in Proposition 1 this upper boundprovides only rather huge bounds for all the exponents. Moreover, the linear formof logarithms (6) is not suitable for applying standard reduction schemes like LLL-reduction. Therefore we give a more detailed account on how to find smaller upperbounds for all the exponents.First, we will prove that all exponents with one possible exception can bebounded in terms of M and M . In particular, Proposition 2 will be useful inthe bound reduction process which we will describe in Section 5. Before we canstate and prove Proposition 2 we prove another helpful lemma: VOLKER ZIEGLER Lemma 9. There exists a permutation σ of { , . . . , } such that α σ (1) ≤ α σ (2) ≤ α σ (3) ≤ α σ (4) ≤ α σ (5) ≤ α σ (6) . and such that one of the following two relations holds: • α σ (1) = α σ (2) = α σ (3) ≤ α σ (4) ≤ α σ (5) ≤ α σ (6) and no two indices out of σ (1) , σ (2) and σ (3) are complementary, • α σ (1) = α σ (2) ≤ α σ (3) = α σ (4) ≤ α σ (5) ≤ α σ (6) and σ (1) and σ (2) arecomplementary.Proof. Let σ be any permutation of { , . . . , } such that α σ (1) ≤ α σ (2) ≤ α σ (3) ≤ α σ (4) ≤ α σ (5) ≤ α σ (6) . Obviously such a permutation exists.Since Lemma 1 we know that α σ (1) = α σ (2) . Assume that σ (1) and σ (2) are notcomplementary. Then there is exactly one S -unit equation out of the system (1)such that this unit equation contains the index σ (1) but not σ (2). Since α σ (1) isminimal it is equal to one exponent of this unit equation, i.e. α σ (1) = α σ (2) = α σ (3) .If σ (1) and σ (3) are complementary then there is exactly one S -unit equationout of the system (1) that does not contain the indices σ (1) and σ (3) but containsthe index σ (2). But since α σ (2) is minimal it is equal to α σ (4) and we have α σ (1) = α σ (2) = α σ (3) = α σ (4) . By exchanging the values of σ (2) and σ (3) the numbers σ (1)and σ (2) are complementary and we have found a permutation that satisfies thesecond case described in the lemma. A similar argument holds if σ (2) and σ (3) arecomplementary. Thus we have proved: If σ (1) and σ (2) are not complementary,then either the first case holds or the second case holds after rearranging the orderof the α ’s, i.e. we have found a suitable permutation σ .Therefore let us assume that σ (1) and σ (2) are complementary. Then there is aunique S -unit equation out of the system (1) such that this unit equation does notcontain the indices σ (1) and σ (2). Thus by Lemma 1 we obtain that α σ (3) = α σ (4) and we are in the second case of the lemma. (cid:3) Similar as in the case of the exponents of p we let τ and ρ be permutations of { , . . . , } such that β τ (1) ≤ β τ (2) ≤ β τ (3) ≤ β τ (4) ≤ β τ (5) ≤ β τ (6) and γ ρ (1) ≤ γ ρ (2) ≤ γ ρ (3) ≤ γ ρ (4) ≤ γ ρ (5) ≤ γ ρ (6) respectively. By exchanging the roles of the α ’s, β ’s and γ ’s we can show thatpermutations τ and ρ with analogous properties as stated in Lemma 9 exist. Letus fix the permutations σ, ρ and τ . We are now in a position to state the nextproposition that will prove to be useful in reducing the huge upper bounds we getfrom Proposition 1. Proposition 2. Assume that log s , log s ≤ M and that log s ≤ M . Moreover,let B p = log p max | β |≤ M / log q | γ |≤ M / log r (cid:8) ν p ( q β r γ − (cid:9) , B q = log q max | α |≤ M / log p | γ |≤ M / log r { ν q ( p α r γ − } ,B r = log r max | α |≤ M / log p | β |≤ M / log q (cid:8) ν r ( p α q β − (cid:9) -DIOPHANTINE QUADRUPLES 9 and B = max { B p , B q , B r } . Then we have α σ (5) ≤ max { M + B p , M } log p , β τ (5) ≤ max { M + B q , M } log q ,γ ρ (5) ≤ max { M + B r , M } log r . In particular, we have max { α σ (5) log p, β τ (5) log q, γ ρ (5) log r } ≤ max { M + B , M } := M . Proof. We give only the details for the proof of the upper bound for α σ (5) since theupper bounds for β τ (5) and γ ρ (5) can be deduced by the same argument.First, let us assume that we are in the first case of Lemma 9. Let us assume forthe moment that |{ , , } ∩ { σ (1) , σ (2) , σ (3) }| ≤ 1. In this case we conclude that |{ , , } ∩ { σ (4) , σ (5) , σ (6) }| ≥ σ (5) ∈ { , , } or σ (6) ∈ { , , } .Thus in any case α σ (5) ≤ M / log p .Now, let us assume that we are again in the first case of Lemma 9 and that |{ , , } ∩ { σ (1) , σ (2) , σ (3) }| = 2. We may assume that σ (3) 6∈ { , , } . If ei-ther σ (5) or σ (6) are contained in { , , } , then we easily conclude that α σ (5) ≤ M / log p . Thus we may assume that σ (4) ∈ { , , } . Further, we deduce that σ (3)and σ (4) are complementary, and without loss of generality we may assume that σ (1) and σ (6) as well as σ (2) and σ (5) are complementary. Let us consider the unitequation s σ (6) s σ (1) − s σ (6) − s σ (1) = s σ (5) s σ (2) − s σ (5) − s σ (2) . Dividing through p α σ (1) and collecting the terms not divisible by p on one side ofthe equation we obtain q β σ (2) r γ σ (2) − q β σ (1) r γ σ (1) = p α σ (5) − α σ (1) × N where N is an integer. Thus we deduce that B log p ≥ B p log p ≥ ν p (cid:0) q β σ (2) − β σ (1) r γ σ (2) − γ σ (1) − (cid:1) ≥ α σ (5) − α σ (1) ≥ α σ (5) − M log p which implies the upper bound for α σ (5) . Note that by our choice that σ (1) , σ (2) ∈{ , , } the exponents satisfy | β σ (2) − β σ (1) | ≤ β and | γ σ (2) − γ σ (1) | ≤ γ .Assume again that we are in the first case of Lemma 9 and that { , , } = { σ (1) , σ (2) , σ (3) } . In this case we can choose σ such that σ (1) and σ (6) as wellas σ (2) and σ (5) are complementary. Now the same line of arguments as in theprevious paragraph yields the same upper bound for α σ (5) .Now, we assume that we are in the second case of Lemma 9. Since σ (1) and σ (2)are complementary exactly one of the two is an element of { , , } . Assume nowthat |{ σ (3) , σ (4) } ∩ { , , }| ≤ 1, then either σ (5) or σ (6) is contained in { , , } and we deduce in any case that α σ (5) ≤ M / log p . Therefore we may assume that σ (3) , σ (4) ∈ { , , } . However σ (1) and σ (2) are complementary and since σ (3)and σ (4) are not we can choose σ such that σ (3) and σ (5) as well as σ (4) and σ (6) are complementary. By the same reasoning as in the proof of the first case wededuce that B log p ≥ B p log p ≥ ν p (cid:0) q β σ (4) − β σ (3) r γ σ (4) − γ σ (3) − (cid:1) ≥ α σ (5) − α σ (3) . Since we are in the second case of Lemma 9 we know that α σ (1) and α σ (2) arecomplementary and therefore { σ (3) , σ (4) , σ (5) , σ (6) } ∩ { , } 6 = ∅ which implies that α σ (3) ≤ M / log p and we obtain the upper bound for α σ (5) alsoin the second case of the proposition. (cid:3) The next lemma shows that if M is large, then the indices of the large exponentsare of a special form. Lemma 10. Let us assume that α σ (5) ≤ M / log p, β τ (5) ≤ M / log q and γ ρ (5) ≤ M / log r . Then either M ≤ M +3 M log(min { p,q,r } ) or { σ (6) , τ (6) , ρ (6) } = { , , } .Proof. Assume that { σ (6) , τ (6) , ρ (6) } 6 = { , , } and let ∗ ∈ { , , } be such that ∗ 6∈ { σ (6) , τ (6) , ρ (6) } . We obviously have that d < s ∗ ≤ p α σ (5) q β τ (5) r γ ρ (5) ≤ exp(3 M )and also c < s = p α q β r γ ≤ exp( M ) . Therefore we obtainmax { p α σ (6) , q β τ (6) , r γ ρ (6) } ≤ s = cd + 1 ≤ exp( M + 3 M ) . Now, taking logarithms yields the content of the lemma. (cid:3) Let us denote by N = { , , , . . . } the set of the non negative integers. Now, wecan provide upper bounds for the largest exponents. Proposition 3. Let us assume that M > M +3 M log(min { p,q,r } ) , σ (6) and x are com-plementary indices and that y, z is another pair of complementary indices, with y ∈ { , , } . If α y = α z < α x (Case I) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , b z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and a y < a x and (cid:18) ≤ b z ≤ M log q , ≤ c z ≤ M or ≤ b z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) , if α x = α z < α y (Case II) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , b z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and a x < a y and (cid:18) ≤ b z ≤ M log q , ≤ c z ≤ M or ≤ b z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) , and if α x = α y (Case III) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , b z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and a x = a y and (cid:18) ≤ b z ≤ M log q , ≤ c z ≤ M or ≤ b z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) . -DIOPHANTINE QUADRUPLES 11 We define C p = max M x,y,z ( ν p q b z r c z (cid:0) p a y q b y r c y − (cid:1) q b y r c y − p a x − a y q b x r c x − !) Case I , C p = max M x,y,z ( ν p q b z r c z (cid:0) p a y q b y r c y − (cid:1) p a y − a x q b y r c y − q b x r c x − !) Case II , C p = max M x,y,z ( ν p q b z r c z (cid:0) p a y q b y r c y − (cid:1) p ( q b y r c y − q b x r c x ) p − !) Case III , where ( · ) p denotes the p -free part. Then α σ (6) ≤ M log p + C p Cases I and II; M + B p log p + C p Case III.Proof. Let us assume that Case I holds. Then we consider the S -unit equation s z s y − s z − s y = s σ (6) s x − s σ (6) − s x which turns into s z ( s y − s y − s x − s σ (6) ( s x − s y − s x . First, we note that ν p ( s y − s x ) = α y = α z and since α x > ν p (cid:18) s σ (6) ( s x − s y − s x (cid:19) = α σ (6) − α y . Note that with these notations and assumptions we have after canceling a commonfactor p α y = p α z the equation s z ( s y − s y − s x − q β z r γ z (cid:0) p α y q β y r γ y − (cid:1) q β y r γ y − p α x q β x r γ x − α x , α y , β x , β y , γ x and γ y satisfying α y < α x and0 ≤ α x log p + β x log q + γ x log r, α y log p + β y log q + γ y log r ≤ M . Moreover due to Lemma 10 we have either 0 ≤ β z ≤ M log q and 0 ≤ γ z ≤ M or0 ≤ β z ≤ M and 0 ≤ γ z ≤ M log r . That is ( α x , β x , γ x , α y , β y , γ y , β z , γ z ) ∈ M x,y,z .Thus we conclude that C p ≥ ν p (cid:18) s z ( s y − s y − s x − (cid:19) = ν p (cid:18) s σ (6) ( s x − s y − s x (cid:19) = α σ (6) − α y . Which implies in view of α y ≤ M / log p and (7) the upper bound for α σ (6) .Almost the same arguments apply in the case that α x = α z < α y holds, i.e. thatCase II holds. We only have to exchanging the roles of x and y .In Case III, that is we have α x = α y , the proof runs along similar arguments.That is we consider the equation s z ( s y − s y − s x − s σ (6) ( s x − s y − s x . However this time we note that ν p ( s y − s x ) = α x + ν p ( q β y − β x r γ y − γ x − ≤ M + B p log p . This implies that α σ (6) ≤ ν p (cid:18) s z ( s y − s y − s x − (cid:19) + ν p ( s y − s x ) , and therefore the upper bound for α σ (6) . Finally let us note that a non vanishingfactor p of s z ( s y − s y − s x would imply that ν p (cid:16) s z ( s y − s y − s x − (cid:17) ≤ 0, thus we may eliminateall possible powers of p in the formula for C p , which accounts in the usage of p -freeparts in the formulas for Case II. (cid:3) Let us note that we can prove similar results for β τ (6) and γ ρ (6) by exchangingthe roles of p, q and r respectively. For the sake of completeness we state the resultsfor β τ (6) and γ ρ (6) in the appendix but do not provide a proof since the proofs areidentical after an appropriate permutation of p, q and r .If we have no explicit bounds for M and M this proposition is hard to apply.However, if we have small, explicit upper bounds for M and M we can estimate C p by applying lower bounds for linear forms in two p -adic logarithms (for detailssee Section 5) which will yield reasonable small bound for M .4. The LLL-reduction In this section we gather some basic facts on LLL-reduced bases, approximationlattices and their applications to Diophantine problems as they can be found in [9,Chapters 5 and 6].Let L ⊆ R k be a k -dimensional lattice with LLL-reduced basis b , . . . , b k and de-note by B be the matrix with columns b , . . . , b k . Moreover, we denote by b ∗ , . . . , b ∗ k the orthogonal basis of R k which we obtain by applying the Gram-Schmidt processto the basis b , . . . , b k . In particular, we have that b ∗ i = b i − i − X j =1 µ i,j b ∗ j , µ i,j = h b i , b j ih b ∗ j , b ∗ j i . Further, let us define l ( L , y ) = ( min x ∈L k x − y k , y 6∈ L min = x ∈L k y k , y ∈ L , where k · k denotes the euclidean norm on R k . It is well known, that by applyingthe LLL-algorithm it is possible to give in polynomial time a lower bound for l ( L , y ) ≥ ˜ c (see e.g. [9, Section 5.4]): Lemma 11. Let y ∈ R k , z = B − y . Furthermore we define • If y 6∈ L let i be the largest index such that z i = 0 and put σ = { z i } ,where {·} denotes the distance to the nearest integer. • If y ∈ L we put σ = 1 .Finally let ˜ c = max ≤ j ≤ k ( k b k k b ∗ j k ) . -DIOPHANTINE QUADRUPLES 13 Then we have l ( L , y ) ≥ ˜ c − σ k b k = ˜ c . In our applications we suppose that we are given real numbers η , η , . . . , η k linearly independent over Q and two positive constants ˜ c , ˜ c such that(8) | η + x η + · · · + x k η k | ≤ ˜ c exp( − ˜ c H ) , where the integers x i are bounded by | x i | ≤ X i with X i given upper bounds for1 ≤ i ≤ k . We write X = max ≤ i ≤ s { X i } . The basic idea in such a situation, dueto de Weger [2], is to approximate the linear form (8) by an approximation lattice.Namely, we consider the lattice L generated by the columns of the matrix A = . . . . . . . . . ⌊ ˜ Cη ⌋ ⌊ ˜ Cη ⌋ . . . ⌊ ˜ Cη k − ⌋ ⌊ ˜ Cη k ⌋ where ˜ C is a large constant usually of the size of about X k . Let us assume thatwe have an LLL-reduced basis b , . . . , b k of L and that we have a lower bound l ( L , y ) > ˜ c with y = (0 , , . . . , −⌊ Cη ⌋ ). Note that ˜ c can be computed by usingthe results of Lemma 11. Then we have with these notations the following lemma(e.g. see [9, Lemma VI.1]): Lemma 12. Assume that S = P k − i =1 X i and T = P ki =1 X i . If ˜ c ≥ T + S ,then inequality (8) implies that we have either x = x = · · · = x k − = 0 and x k = − ⌊ ˜ Cη ⌋ ) ⌊ ˜ Cη k ⌋ ) or (9) H ≤ c (cid:18) log( ˜ C ˜ c ) − log (cid:18)q ˜ c − S − T (cid:19)(cid:19) . Reduction of the bounds In this section we describe how we can reduce the rather huge bounds for theexponents α i , β i , γ i for i = 1 , . . . , S = { p, q, r } andhow it is in practice possible to enumerate for the given set S all S -Diophantinequadruples. We give the full details for the proof of Theorem 1, i.e. the case that S = { , , } and will outline the strategy how to find all S -Diophantine quadruplesfor a general set of three primes S . The reduction process follows in eight steps: Step I: We compute an upper bound for M by using Proposition 1. In the casethat S = { , , } we obtain M < . · . Note that in the case that P < M < . · . Step II: We apply the LLL-reduction method explained in Section 4 to thelinear forms of logarithms (3) and (5) in order to obtain a small bound for M .In particular, we apply Lemma 12 to the linear form (3). In the case that S = { , , } we put k = 3 and η = log 5 , η = log 3 and η = log 2. Moreoverwe have 0 ≤ A ′ , B ′ , C ′ ≤ M , i.e. we have to put X = X = X = 2 . · .Moreover let us put ˜ c = 2 and ˜ c = 1. With ˜ C = 10 the assumptions of Lemma12 are satisfied and we obtain2 exp( − . > | A ′ log 2 + B ′ log 3 + C ′ log 5 | > p α q β r γ which implies log s ≤ . 4. Since the exact same computations also apply to (5)we also obtain that log s < . M < . s < s s < exp(272 . M ≤ . Step III: By a direct computation we are able to compute B p , B q and B r by abrute force algorithm. The search implemented in Sage [14] takes on a usual PC afew seconds. In particular we obtain in the case that S = { , , } the bounds B p log p = 18 , B q log q = 13 , B r log r = 8 . By Proposition 2 this yields α σ (5) ≤ , β τ (5) ≤ , γ ρ (5) ≤ . In particular, we have M = M ≤ . Step IV: In this step we use the theorem of Bugeaud and Laurent (Lemma 5)to estimate the quantities C p and obtain a new upper bound for α σ (6) due to Propo-sition 3 (and the upper bounds for C q and C r by Propositions 5 and 6 in the Ap-pendix). Since the computations for C q and C r are similar we give the details onlyfor C p for Case I note that the upper bounds for Cases II and III can be deducedsimilarly.To find an upper bound for C p we split up the set M x,y,z into two subsets M ′ x,y,z , M ′′ x,y,z ⊆ M x,y,z with M ′ x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , b z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and a y < a x and 0 ≤ b z ≤ M log q , ≤ c z ≤ M (cid:27) and M ′′ x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , b z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and a y < a x and 0 ≤ b z ≤ M, ≤ c z ≤ M log r (cid:27) and also consider the two quantities C ′ p = max M ′ x,y,z ( ν p q b z r c z (cid:0) p a y q b y r c y − (cid:1) q b y r c y − p a x − a y q b x r c x − !) and C ′′ p = max M ′′ x,y,z ( ν p q b z r c z (cid:0) p a y q b y r c y − (cid:1) q b y r c y − p a x − a y q b x r c x − !) . Of course we have C p = max {C ′ p , C ′′ p } and we have to find upper bounds for C ′ p (seeStep IV (a)) and C ′′ p (see Step IV (b)). Similarly we can define C ′ q , C ′′ q , C ′ r and C ′′ r which will yield bounds for C q and C r respectively. Step IV (a): In this step we compute an upper bound for C ′ p . We put η = r, η = q b z ( p a y q b y r c y − q b y r c y − p a x − a y q b x r c x -DIOPHANTINE QUADRUPLES 15 and b = c z and b = 1. We find that h ( η ) ≤ max { b z log q + log( p a y q b y r c y − , log( q b y r c y − p a x − a y q b x r c x ) }≤ M + M since we have 0 ≤ b z ≤ M log q and0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M . Applying Lemma 5 with this data we obtain a new hopefully smaller upper boundfor A = α σ (6) . Step IV (b): In this step we compute an upper bound for C ′′ p . In this case weput η = q , η = r c z ( p a y q b y r c y − q b y r c y − p a x − a y q b x r c x and b = b z and b = 1. We find that h ( η ) ≤ max { c z log r + log( s y − , log( s y − s x ) } ≤ M + M since we have 0 ≤ c z ≤ M log r and0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M . Again we apply Lemma 5 with this data and we obtain an upper bound for A = α σ (6) in this case.Similarly we find upper bounds in Cases II and III. We also find upper boundsfor B = β τ (6) and C = γ ρ (6) by using Propositions 5 and 6 instead of Propositions3. In the case that S = { , , } we obtain A < . · , B < . · , C < . · . Step V: We repeat the process of the Steps II-IV iteratively with the new foundbounds for M and M four more times and will find significantly smaller bounds.In the case that S = { , , } we obtain after five LLL-reductions the upper bounds M < . M = M < . 248 and A < . · , B < . · , C < . · . We implemented the Steps I-V in Sage. It took a usual desk top PC only a fewseconds to obtain the bounds stated in Step V. Before we proceed with Step VI letus summarize what we proved so far in the case that S = { , , } : Proposition 4. Assume that ( a, b, c, d ) is a { , , } -Diophantine Quadruple with a < b < c < d . Then log s , log s < . and log s < . . Step VI: In this step we find all { p, q, r } -Diophantine triples ( a, b, c ) such thatlog( ab + 1) = log s , log( ac + 1) = log s < M and log( bc + 1) = log s < M .Therefore we proceed as follows:(1) We enumerate all S -units s = p α q β r γ with log s < M .(2) We enumerate all S -units s = p α q β r γ with log s < min { log s , M } and s ∤ s . In particular, we consider only those s which satisfy γ > γ in case that α ≤ α and β ≤ β . (3) We compute for all pairs s , s the quantity B c = s − s gcd( s ,s ) . Due toLemma 2 we have c ≤ B c . If ( B c − B c + 1 < s we discard the pair s , s since otherwise we would have bc + 1 < ( B c − B c + 1 < s = bc + 1 , a contradiction.(4) We enumerate all S -units s = p α q β r γ with s < s such that • if p = 2 and α , α > 1, then α ≤ • if p ≡ α , α > 0, then α = 0; • if q ≡ β , β > 0, then β = 0; • if r ≡ γ , γ > 0, then γ = 0.That is we enumerate all S -units s such that the content of Lemma 3 isnot violated for the S -Diophantine triple ( a, b, c ).(5) For all these triples ( s , s , s ) we compute a (cid:3) = ( s − s − s − . We discardall triples ( s , s , s ) for which a (cid:3) is not the square of an integer. For theremaining triples we compute a = √ a (cid:3) .(6) We discard all triples ( s , s , s ) for which b = s − a and c = s − a are notintegers.(7) We store all triples ( a, b, c ) in a list Triples .In the case that S = { , , } the list Triples consists of the S -Diophantinetriples { (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (3 , , , (1 , , , (2 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , } . Let us note that Step VI is the bottle neck of our algorithm. It took more thansix and a half hours to find all these triples using a usual PC (Intel i7-8700) on asingle core. Step VII: Let us fix one S -Diophantine triple ( a, b, c ) from the list Triples computed in Step VI. Let us assume that this triple can be extended to a S -Diophantine quadruple. In this step we reduce the previously found huge upperbound for s = ad + 1 by using the LLL-reduction (Lemma 12). Therefore weconsider the inequality ba · ad + 1 bd + 1 − b − aabd + a < ad + 1and taking logarithms we obtain(10) | log( b/a ) + ( α − α ) log p + ( β − β ) log q + ( γ − γ ) log r | < s We apply Lemma 12 to this Diophantine inequality that is we have k = 3 , η = log r, η = log q, η = log p, η = log( b/a ) . Moreover we put X = A , X = B and X = C , where A, B and C are the boundsfor α σ (6) , β τ (6) and γ ρ (6) found in Step V. We distinguish now between two cases. -DIOPHANTINE QUADRUPLES 17 Case I: Let us assume that p, q, r and a/b are not multiplicatively dependent.For example this happens for the { , , } -Diophantine triple (1 , , l ( L , y ) by using Lemma 11 with A = ⌊ C log r ⌋ ⌊ C log q ⌋ ⌊ C log p ⌋ = ⌊ ˜ C log 5 ⌋ ⌊ ˜ C log 3 ⌋ ⌊ ˜ C log 2 ⌋ and y = (0 , , −⌊− ˜ C log( b/a ) ⌋ ) T = (0 , , −⌊− ˜ C log 7 ⌋ ) T . With ˜ C = 10 we obtain by Lemma 12 the bound log s < . { , , } -Diophantine triple (1 , , Case II: Let us assume that p, q, r and a/b are multiplicatively dependent andthat b/a = p x p q x q r x r . Then Inequality (10) turns into | ( α − α + x p ) log p + ( β − β + x q ) log q + ( γ − γ + x r ) log r | < s . Thus we compute l ( L , y ) by using Lemma 11 with A = ⌊ C log r ⌋ ⌊ C log q ⌋ ⌊ C log p ⌋ and y = (0 , , T . We put X = A + | x p | , X = B + | x q | and X = C + | x r | and ˜ C sufficiently large and obtain an upper bound for log s . For example in the case ofthe { , , } -Diophantine triple (1 , , 5) we have x = x = 0 and x = 1 and obtainwith this strategy log s < . C = 10 .Let us note that in the case that S = { , , } we obtain in all cases the upperbound log s < Step VIII: Let M denote the upper bound for log s found in Step VII. Weenumerate all ad + 1 = s = p α q β r γ with log s < M and for each triple ( a, b, c )found in Step VI we compute d = s − a . We discard all quadruples ( a, b, c, d ) forwhich d is not an integer and for which bd + 1 and cd + 1 are not S -units.Note that in the case that S = { , , } we have M = 40. However a ratherquick computer search yields that no triple can be extended to a quadruple and theproof of Theorem 1 is therefore complete. Remark . Finally let us note that the algorithm presented in this paper would alsowork for arbitrary S with | S | ≥ p -adic logarithms (e.g.the results from [15]) instead of the results due to Bugeaud and Laurent [1] onlinear forms in two p -adic logarithms. This will lead to much larger bounds for M and M than what we get in the case | S | = 3.Moreover, Step VI in which we search for extendable triples seems to get unfea-sible in the case | S | ≥ 4. Thus all together a systematic search for S -Diophantinequadruples or even quintuples in the case | S | ≥ Appendix – Bounds for β τ (6) and γ ρ (6) In this appendix we explicitly state the upper bounds for β τ (6) and γ ρ (6) we getby adjusting the proof of Proposition 3. A bound for β τ (6) is given by: Proposition 5. Let us assume that M > M +3 M log(min { p,q,r } ) , τ (6) and x are com-plementary indices and that y, z is another pair of complementary indices, with y ∈ { , , } . If β y = β z < β x (Case I) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , a z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and b y < b x and (cid:18) ≤ a z ≤ M log p , ≤ c z ≤ M or ≤ a z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) , if β x = β z < β y (Case II) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , a z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and b x < b y and (cid:18) ≤ a z ≤ M log p , ≤ c z ≤ M or ≤ a z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) , and if β x = β y (Case III) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , a z , c z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and b x = b y and (cid:18) ≤ a z ≤ M log p , ≤ c z ≤ M or ≤ a z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) . We define C q = max M x,y,z ( ν q p a z r c z (cid:0) p a y q b y r c y − (cid:1) p a y r c y − p a x q b x − b y r c x − !) Case I , C q = max M x,y,z ( ν q p a z r c z (cid:0) p a y q b y r c y − (cid:1) p a y q b y − b x r c y − p a x r c x − !) Case II , C q = max M x,y,z ( ν q p a z r c z (cid:0) p a y q b y r c y − (cid:1) q ( p a y r c y − p a x r c x ) q − !) Case III , where ( · ) q denotes the q -free part. Then β τ (6) ≤ M log q + C q Cases I and II; M + B q log q + C q Case III. A bound for γ ρ (6) is given by: Proposition 6. Let us assume that M > M +3 M log(min { p,q,r } ) , ρ (6) and x are com-plementary indices and that y, z is another pair of complementary indices, with -DIOPHANTINE QUADRUPLES 19 y ∈ { , , } . If γ y = γ z < γ x (Case I) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , a z , b z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and c y < c x and (cid:18) ≤ a z ≤ M log p , ≤ b z ≤ M or ≤ a z ≤ M, ≤ b z ≤ M log q (cid:19)(cid:27) , if α x = α z < α y (Case II) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , a z , b z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and c x < c y and (cid:18) ≤ a z ≤ M log p , ≤ c z ≤ M or ≤ a z ≤ M, ≤ c z ≤ M log r (cid:19)(cid:27) , and if α x = α y (Case III) we put M x,y,z = (cid:26) ( a x , b x , c x , a y , b y , c y , a z , b z ) ∈ N :0 ≤ a x log p + b x log q + c x log r, a y log p + b y log q + c y log r ≤ M , and c x = c y and (cid:18) ≤ a z ≤ M log p , ≤ b z ≤ M or ≤ a z ≤ M, ≤ b z ≤ M log q (cid:19)(cid:27) . We define C r = max M x,y,z ( ν r p a z q b z (cid:0) p a y q b y r c y − (cid:1) p a y q b y − p a x q b x r c x − c y − !) Case I , C r = max M x,y,z ( ν r p a z q b z (cid:0) p a y q b y r c y − (cid:1) p a y q b y r c y − c x − p a x q b x − !) Case II , C r = max M x,y,z ( ν r p a z q b z (cid:0) p a y q b y r c y − (cid:1) r ( p a y q b y − p a x q b x ) r − !) Case III , where ( · ) r denotes the r -free part. Then γ ρ (6) ≤ M log r + C r Cases I and II; M + B r log r + C r Case III. References [1] Y. Bugeaud and M. Laurent. Minoration effective de la distance p -adique entre puissances denombres alg´ebriques. J. Number Theory , 61(2):311–342, 1996.[2] B. M. M. de Weger. Solving exponential Diophantine equations using lattice basis reductionalgorithms. J. Number Theory , 26(3):325–367, 1987.[3] P. Erd¨os, C. L. Stewart, and R. Tijdeman. Some Diophantine equations with many solutions. Compositio Math. , 66(1):37–56, 1988.[4] P. Erdos and P. Turan. 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