SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS ANDREJ DUJELLA AND ANA JURASIĆ Abstract. In this paper we give some new examples of polynomial D(n)- triples and quadruples i.e. sets of polynomials with integer coefficients such that the product of any two of them plus a polynomial n Z[X] is a square of a polynomial with integer coefficients. The examples illustrate various theoretical properties and constructions for a quadratic polynomial n which appeared in recent papers. One of the examples gives a partial answer to the question about number of distinct D(n)-quadruples if n is an integer that is the product of twin primes. 000 Mathematics Subject Classification: 11C08 11D99. Keywords: Diophantine m-tuples polynomials. 1. Introduction Let n be a nonzero integer. A set {a 1 a... a m } of m distinct positive integers is called a Diophantine m-tuple with the property D(n) or simply D(n)-m-tuple if a i a j + n is a perfect square for all 1 i < j m. An improper D(n)-m-tuple is an m-tuple with the same property but with relaxed condition that the elements a i can be integers and need not be distinct. Diophantus found the quadruple {1 68 105} with the property D(56). The first Diophantine quadruple with the property D(1) the set {1 8 10} was found by Fermat (see [4]). Diophantine triples and quadruples with the property D(n) can be classified as regular or irregular depending on whether they satisfy the conditions given in next two definitions. Let {a b c} be a D(n)-triple and (1) where r s t are positive integers. ab + n = r ac + n = s bc + n = t Definition 1. A D(n)-triple {a b c} is called regular if it satisfies the condition () (c b a) = 4(ab + n). Equation () is symmetric under permutations of a b c. Also from () using (1) we get c ± = a + b ± r and we have ac ± + n = (a ± r) bc ± + n = (b ± r). Definition. A D(n)-quadruple {a b c d} is called regular if it satisfies the condition () n(d + c a b) = 4(ab + n)(cd + n). 1

ANDREJ DUJELLA AND ANA JURASIĆ Equation () is symmetric under permutations of a b c d. Since the right hand side of () is a square it follows that a regular D(n)-quadruple can exist only if n is a square whereas regular D(n)-triples exist for all n. When n is a square the quadruple formed by adding a zero to a regular triple is an improper regular quadruple. Equation () is a quadratic equation in d with roots and we have d ± = a + b + c + (abc ± rst) n ad ± + n = 1 n (rs ± at) bd ± + n = 1 n (rt ± bs) cd ± + n = 1 n (st ± cr). An irregular D(n)-tuple is one that is not regular. A semi-regular D(n)-quadruple is one which contains a regular triple and a twice semi-regular D(n)-quadruple is one that contains two regular triples. For n = 1 we have the following conjecture (see [10]). Conjecture 1. If {a b c d} is a D(1)-quadruple and d > max{a b c} then d = d +. It is clear that Conjecture 1 implies that there does not exist a D(1)-quintuple. Baker and Davenport [1] proved Conjecture 1 for the triple {a b c} = {1 8} with the unique extension d = 10. Dujella and Pethő [16] proved it for all triples of the form {1 c}. Conjecture 1 was recently verified for all triples of the form {k 1 k +1 c} (see [7 18 ]). Dujella [10] proved that there are only finitely many D(1)-quintuples and there does not exist a D(1)-sextuple. Fujita [19] proved that any D(1)-quintuple contains a regular Diophantine quadruple i.e. if {a b c d e} is a D(1)-quintuple and a < b < c < d < e then d = d +. Polynomial variant of the above problem was first studied by Jones [ ] and it was for the case n = 1. Definition. Let n Z[X] and let {a 1 a... a m } be a set of m different nonzero polynomials with integer coefficients. We assume that there does not exist a polynomial p Z[X] such that a1 p... am p and n p are integers. The set {a 1 a... a m } is called a polynomial D(n)-m-tuple if for all 1 i < j m the following holds: a i a j + n = b ij where b ij Z[X]. The assumption concerning the polynomial p means that not all elements of {a 1 a... a m } are allowed to be constant. Since the constructions from Definitions 1 and are obtained using only algebraic manipulations they are also valid in the polynomial case. Dujella and Fuchs [11] proved that all polynomial D(1)-quadruples in Z[X] are regular. Note that this is not true in C[X] as Dujella and Jurasić showed in [15] by giving the following example { (X 1) + X + + X + } of a D(1)-quadruple in Q( )[X] which is not regular. In [15] it is shown that there does not exist a Diophantine 8-tuple in C[X] (see also [1]). For polynomial D(n)-m-tuples with n constant from [15 Theorem 1] it follows that m 7 for all n Z\{0}. Dujella and Fuchs jointly with Tichy [1] and later

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS with Walsh [14] considered the case n = µ 1 X + µ 0 with integers µ 1 0 and µ 0. Let L = sup{ S : S is a polynomial D(µ 1 X +µ 0 )-m-tuple for some µ 1 0 µ 0 Z} where S denotes the number of elements in the set S and let L k be the number of polynomials of degree k in a polynomial D(µ 1 X + µ 0 )-m-tuple S. The results from [14] are sharp bounds L 0 1 L 1 4 L k for all k and finally L 1. Jurasić [4] considered the case where n is a quadratic polynomial in Z[X]. Let Q = sup{ S : S is a polynomial D(µ X + µ 1 X + µ 0 )-m-tuple for some µ 0 µ 1 µ 0 Z} and let Q k be the number of polynomials of degree k in a polynomial D(µ X + µ 1 X + µ 0 )-m-tuple S. In [4] it was proved that Q 98 and that if a polynomial D(n)-m-tuple for a quadratic n contains only polynomials of odd degree then m 18. Moreover Q 0 Q 1 4 Q 81 Q 5 Q 4 6 and Q k for k 5. The bounds for Q 0 Q 1 and Q k for all k 5 are sharp. In Section we give some examples of a polynomial D(n)-triples and quadruples where n is a quadratic polynomial. These examples show that several auxiliary results from [4] are sharp i.e. in the cases when the existence of Diophantine triples with certain properties cannot be excluded such triples indeed exist. In Section we study D(n)-quadruples if n is an integer that is the product of twin primes and we construct a polynomial Diophantine quadruple for the quadratic polynomial corresponding to that case. In Section 4 we give polynomial examples for all cases of Gibbs construction of adjugates of a D(n)-quadruples. In some of them we use quadruples constructed in Sections and.. Diophantine triples with elements of the same degree Although the final bound from [4 Theorem 1] Q 98 is likely very far from being optimal the bounds from [4] for polynomials of fixed degree i.e. bounds for Q k are sharp or almost sharp except for k =. In particular they are sharp for k = 0 k = 1 and k 5 in view of the following examples given in [4]: the set { 5} is a polynomial D(9X + 4X + 1)-pair the set {X X + 8 X + 5X + 0} is a polynomial D( (X + 9)(X 1))-quadruple the set {X l 1 + X X l 1 + X l + X 4X l 1 + 4X l + 5X} is a polynomial D( X )-triple for any integer l and the set {X l + X l X l + X l + 4X 4X l + 4X l + 8X} is a polynomial D(4X )-triple for any integer l 1. However even in the cases k = 4 several auxiliary results from [4] are (almost) sharp as it will be illustrated by concrete examples in this section. The main role in the proofs of the upper bounds on Q k k has [1 Lemma 1]. It says that for a polynomial D(n)-triple {a b c} for which (1) holds there exist polynomials e u v w Z[X] such that ae + n = u be + n = v ce + n = w. More precisely (4) e = n(a + b + c) + abc rst

4 ANDREJ DUJELLA AND ANA JURASIĆ u = at rs v = bs rt w = cr st and (5) We define (6) and we easily get (7) Moreover we have (see [4]) (8) (9) (10) c = a + b + e n + (abe + ruv). n e = n(a + b + c) + abc + rst e e = n (c a b r)(c a b + r). e = n(a + b c) + rw e = n(a b + c) + sv e = n( a + b + c) + tu. Let Z + [X] denote the set of all polynomials with integer coefficients with positive leading coefficient. For a b Z[X] a < b means that b a Z + [X]. The polynomial e had an important role in getting upper bounds for Q k k and also in classifying various types of Diophantine triples with the same degree. In [4] all possibilities for e for a given quadratic polynomial n were investigated. Here we are interested to find examples of D(n)-triples which illustrate these possibilities. Trivial case is e = 0 for which we have a regular triple {a b a + b + r}. Let {a b c} where a < b < c be a polynomial D(n)-triple containing only quadratic polynomials and let e Z[X] be defined by (4). We are looking at extensions of {a b} to a polynomial D(n)-triple {a b c} with c > b and then at the corresponding e Z[X] defined by (4). We have (see [4 Lemmas 4-8]): A) If n and e have a common linear factor but n e (we consider divisibility over Q[X]) then n = n 1 n where n 1 n are linear polynomials over Q such that n 1 n. For fixed a and b there exist at most two such e-s. B) If n e then n = λn 1 where λ Q\{0} and n 1 is a linear polynomial over Q. For fixed a and b there is at most one such e. C) If e Z \ {0} then for fixed a and b there is at most one such e. D) If e is a linear polynomial which does not divide n then for fixed a and b there is at most one such e. E) Let e be a quadratic polynomial which does not have a common nonconstant factor with n. Then there is at most one polynomial c c such that {a b c } a < b < c is a polynomial D(n)-triple and f Z[X] obtained by applying (4) on that triple (f = n(a + b + c ) + abc rs t where ac + n = s bc + n = t ) is a quadratic polynomial which does not have a common nonconstant factor with n or the analogous statement holds for {a b c} a < b < c b b or for {a b c} a < b < c a a (see [4 Lemma 8] for details). In [4 Lemmas 9-11] the following results were proved for the sets of polynomials of degree k. Let {a b c} a < b < c be a polynomial D(n)-triple of polynomials of degree k then beside e = 0 we have the following possibilities. If k = : A) For fixed a and b there is at most one e Z\{0} defined by (4). B) Let e Z[X] be a linear polynomial defined by (4). Then e n. For fixed a and b there exist at most two such e-s. For k = 4 we have:

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS 5 A4) Let e Z\{0} be defined by (4). Then for fixed a and b there exist at most three such e-s. For k 5 we have only trivial situation e = 0 (see [4 Proposition 4.)]). The proofs of [4 Lemmas 4-8] give us methods for a construction of the examples for the cases where k =. The first polynomial D(n)-triple is an example for the case A) obtained using [4 Lemma 5]. Let n = n 1 n and e = e 1 e where n i e i for i = 1 are linear polynomials over Q. In the proof of [4 Lemma 4] we have a = A (X ϕ 1 )(X ϕ ) where A N and ϕ 1 ϕ Q. Analogously b = B (X χ 1 )(X χ ) c = C (X λ 1 )(X λ ) where B C N and χ i λ i Q for i = 1. From [1 Lemma 1] we obtain bν + n = v1 where v = n 1 v 1 and cν + n = w1 where w = n 1 w 1 and v 1 w 1 Q[X] deg(v 1 ) deg(w 1 ) 1 ν Q\{0}. Assume that at least one of the polynomials u 1 v 1 and w 1 has a degree equal to 0. Since deg(e) = deg(n) = deg(r) = if deg(w) = 1 then by (8) deg(a + b c) = 1. Hence A + B = C. Analogously if deg(v) = 1 then by (9) we obtain C = A + B < B a contradiction. Also if deg(u) = 1 then by (10) we get C = A B < A a contradiction. Therefore only w 1 can have a degree equal to 0. From [4 Lemma 4] we have n = ε ε 1 X + ε 1ϕ 1 ε ϕ u 1 = ε 1 + ε X ε 1ϕ 1 + ε ϕ where A ν = ε 1 ε and ε 1 ε Q\{0}. Analogously for b and c we get n = τ τ 1 X + τ1χ1 τχ v 1 = τ1+τ X τ1χ1+τχ and n = ψ ψ1 X + ψ1λ1 ψλ w 1 = ψ1+ψ X ψ1λ1+ψλ where B ν = τ 1 τ C ν = ψ 1 ψ and τ i ψ i Q\{0} for i = 1. We conclude that ε ε 1 = τ τ 1 = ψ ψ 1 := η and ψ 1 + ψ = 0. Hence ψ 1 = η ψ = η so C ν = η 4. Note that ν = Q for some Q Q\{0}. Multiplying A +B = C by ν we get ε 1 (ε 1 +η)+τ 1 (τ 1 +η) = η 4 from which we obtain ε 1 = η ±BQ and τ 1 = η ±AQ. If we choose for example A = B = 4 Q = 1 η = 10 ϕ = 0 χ 1 = 1 χ = using the equations ε 1 ϕ 1 ε ϕ = τ 1 χ 1 τ χ ab + n = r and (5) we obtain a polynomial D(6(5X 6)(9X 54))-triple {9X(X 1) 16(X 1)(X ) (5X + 14)(5X 6)}. Here e = 196(9X 54). Similarly we find the example for the case B) described in [4 Lemma 4]. From (7) it follows that n ē and from (4) and (6) we get ē e = 4rst so n rst. Therefore by (4) n abc. Since in this case n = λn 1 where λ Q\{0} and n 1 is a linear polynomial over Q we may assume that n 1 a. Now from [1 Lemma 1] we get that n 1 u 1 where u 1 Q[X] and u = n 1 u 1. Hence n 1 u 1 so n 1 a. For A = B = 4 Q = 1 η = 10 ϕ 1 = ϕ = 0 χ = 1 analogously as in the previous case we obtain a polynomial D(900X )-triple {9X 16(X + 4)(X 1) 5(X )(X + )}. Here e = 400X. This triple can be extended to a regular D(n)-quadruple {9X 16(X + 4)(X 1) 5(X )(X + ) d}

6 ANDREJ DUJELLA AND ANA JURASIĆ in two ways. From () we obtain d = 6 and d = 4(X +1)(X )(X +)(X +1). For the case C) we follow the strategy from the proof of [4 Lemma 6]. Let a = A (X + α 1 X + α 0 ) b = B (X + β 1 X + β 0 ) and c = C (X + γ 1 X + γ 0 ) where α i β i γ i Q for i = 0 1. By [1 Lemma 1] there exist u v w Z[X] such that ae + n = u be + n = v and ce + n = w. Here we have deg(u) = deg(v) = deg(w) = and the leading coefficients of the polynomials u v w are equal to the leading coefficient of n up to a sign. Denote by µ ϑ ν ω the leading coefficients of n u v w respectively. By (8) ω = µ by (10) ϑ = µ and by (9) ν = µ. From ae + n = u it follows that (11) u + n = ε 1 u n = ε (X + α 1 X + α 0 ) where ε 1 ε = A e and ε 1 ε Q\{0}. From (11) we obtain n = ε X + ε α 1 X + ε α 0 ε 1 u = ε X + ε α 1 X + ε α 0 + ε 1. For b and c we have equations analogous to (11). Let B e = τ 1 τ and C e = ψ 1 ψ where τ i ψ i Q\{0} for i = 1. It follows that n = τ X + τ β 1 X + τ β 0 τ 1 v = τ X + τ β 1 X + τ β 0 +τ 1 and n = ψ X + ψ γ 1 X + ψ γ 0 ψ 1 w = ψ X + ψ γ 1 X + ψ γ 0 +ψ 1. From the expressions for n we conclude that ε = τ = ψ and α 1 = β 1 = γ 1. Using this and the fact that the discriminants of the polynomials ae+n be+n and ce+n must be equal to 0 we obtain a system of equations with infinitely many solutions. One of that solutions is A = B = 4 e = (α 0 β 0 ) ε = ε 1 = α 0 + β 0 γ = 8β 0 5α 0. The equations (1) are satisfied for a = 4X b = 16X c = 6X 1 n = 4X + 1. Hence {4X 16X 6X 1} is a polynomial D(4X +1)-quadruple where e = 4. This triple can be extended to a twice semi-regular polynomial D(4X + 1)-quadruple (1) {4X 16X 6X 1 4X 4}. For the triple {4X 6X 1 4X 4} we have e = 180X 0. For a = 4X and b = 16X we also have e = 640X 40 for which we get a polynomial D(4X + 1)-triple {4X 16X 4X 4}. For e = 0 we have a regular polynomial D(4X + 1)-triple {4X 16X 6X }. Observe that quadruple (1) can be obtained by taking k = 1 and m = 4X in [5 formula (15)]. Let us now find an example for the case D) using the proof of [4 Lemma 7]. Using (8) we obtain (1) 1 rξ = σn where ξ σ Q\{0}. We further obtain e = 1 ξ (a + b + r nξ). + σ Since e must be a linear polynomial from the coefficient with X on the right side of the previous equation we conclude that ξ = (A+B) µ where µ is a leading

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS 7 coefficient of n. Now from (1) it follows that σ = AB(A+B) µ. For a = X + X B = and µ = 6 using the relation ab + n = r and (5) we obtain a polynomial D(6X + X + 1)-triple {X(X + 1) (X 1)(X + 1) 9X 9X }. Here e = 4X 8. Notice that for a = X(X + 1) and b = (X 1)(X + 1) from e = 1 ξ +σ (a + b r + nξ) we obtain e = 8(5X + )(4X 1) and then c = 81X + 7X 15. We will now illustrate the case E) described in [4 Lemma 8 1)]. Let a = X b = 16X + 8 and n = 16X + 9. We obtain ψ = ϕ = 1 and then e = 7X + 16 f = X + 18. For e we get c = 100X + 44 and for f we get c = 6X + 0 so we have two polynomial D(n)-triples {a b c} and {a b c }. The second of them is a part of a twice semi-regular polynomial D(n)-quadruple {X 16X + 8 5X + 14 6X + 0} (this quadruple can be obtained by taking k = X in the formula for a D(16k + 9)-quadruple from [5 Section 6]). For the cases where k = and k = 4 we find the examples using a method of undetermined coefficients and then fixing some of the coefficients to 0 or to 1. Let us first find the example for the case A). Let a = X + α X + α 1 X + α 0 n = µ X +µ 0 u = µ X +ϑ 1 X +ϑ 0 be a polynomials over Q. From ae+n = u we obtain e = ϑ 1 µ α 0 = ϑ 0 µ 0 α 1 = ϑ 0 and α = µ 0µ ϑ 0 µ ϑ 1. ϑ 1 µ µ ϑ 1 µ Let b = X + β X + β 1 X + β 0 and v = µ X + ν 1 X + ν 0 be a polynomials over Q. From be + n = v and the obtained expression for e we get ϑ 1 = ν 1. Also β = µ 0µ ν 0 µ ϑ 1 β 1 = ν 0 and β 0 = ν 0 µ 0. We now equate the ϑ 1 µ µ ϑ 1 µ expression (X + α X + α 1 X + α 0 )(X + β X + β 1 X + β 0 ) + µ X + µ 0 with the square of a polynomial of degree. It is easy to obtain equal coefficients of X 6 X 5 X 4 X at both sides of the equation. Then comparing the coefficients of X X 1 and X 0 we get three equations in unknowns µ µ 0 ϑ 1 ϑ 0 ν 0. If µ 0 = 0 the system has a nontrivial rational solution for ν 0 = 7τ 6 τ Q. Namely µ = 9τ 4 ϑ 1 = 9τ 5 ϑ 0 = 9τ 6. By taking τ = 1 and multiplying the elements of the obtained rational triple by we finally get a polynomial D(6X )-triple {(X 1)(X + 1)(X 1) (X + )(X + X + ) 4(X + )(X + X + )} with e = 196. We can extend this triple to the regular quadruple in Q[X] by taking the fourth element d = 4 9 X (X + )(X + 5)(X + 1)(X + X + 4). Let us consider the case B). polynomials over Q in the form If we write ae + n = u where a e n u are (X + α X + α 1 X + α 0 )((ε φ )X + γ) + (φx ) = (εx + ϑ 1 X + ϑ 0 ) the coefficients of X 4 are equal on both sides of the equation. Comparing the other coefficients we obtain α = γ ϑ 1ε ε + φ α 1 = ϑ 1ε ϑ 1φ + γ γϑ 1 ε φε + φ + ϑ 0 ε ϑ 0 εφ ( ε + φ )

8 ANDREJ DUJELLA AND ANA JURASIĆ α 0 = γϑ 1 ε +γϑ 1 φ γ +γ ϑ 1ε+γφε γφ γϑ 0ε +γϑ 0εφ +ϑ 0ϑ 1ε 4 4ϑ 0ϑ 1ε φ +ϑ 0ϑ 1φ 4 ( ε +φ ) ϑ 1 = ϑ 0ε φ ϑ 0 εφ +γ +ϑ 0 ε ϑ 0 φ ε φε +εφ +φ (φ+ε)γ. Similarly from be + n = v we get β = γ ν 1ε ε + φ β 1 = ν 1ε ν1φ + γ γν 1 ε φε + φ + ν 0 ε ν 0 εφ ( ε + φ ) β 0 = γν 1 ε +γν 1 φ γ +γ ν 1ε+γφε γφ γν 0ε +γν 0εφ +ν 0ν 1ε 4 4ν 0ν 1ε φ +ν 0ν 1φ 4 ( ε +φ ) ν 1 = ν 0ε φ ν 0 εφ +γ +ν 0 ε ν 0 φ ε φε +εφ +φ (φ+ε)γ where v = εx + ν 1 X + ν 0 and b = X + β X + β 1 X + β 0 are polynomials over Q. We are left with the condition that (X + α X + α 1 X + α 0 )(X + β X + β 1 X + β 0 ) + φx is equal to a square of a polynomial r with degree. It is easy to obtain the equal coefficients of X 6 X 5 X 4 X on both sides of the equation. Comparing the coefficients of X X 1 X 0 on both sides of the equation we get three equations in the remaining γ unknowns. For ν 0 = ε(φ + ε) and ϑ 0 = γ (φ + ε) (φ + ε) εφ the equations for X1 and X 0 are satisfied. Inserting those expressions for ν 0 and ϑ 0 into the equation for X we obtain (14) ε φ 9 + 6φ 8 ε + 15ε 4 φ 7 + 0ε 5 φ 6 + 15ε 6 φ 5 + 6ε 7 φ 4 + ε 8 φ 4γ 4 φ 16γ 4 εφ 16ε γ 4 = 0. One solution of the equation (14) is ε = φ = 9 γ = 54 for which we have a = X + 5 X 5X + b = X X 9X + 7 n = 9X r = X + 1 X 9X + 9 e = 7X + 54 u = X 1X + 9 v = X + 7X 7 and c = 4 9 X + 10 9 X + 16 9 X +. Multiplying those expressions by 18 we get integer coefficients. Indeed {9(X 1)(X + 7X ) 9(X + )(X )(X ) 4(X + 1)(X + X + )} is a polynomial D(916X )-triple with e = 419904X + 1498. Extending this triple to the regular quadruple in Q[X] we get the fourth element d = 4 9 X (X + )(X 5)(X + 7)(X + X 1). Let us take a closer look at equation (14). It leads to the condition 4φ ε (φ + ε) 6 (φ + ε) = U 4. By substituting φ = T ε = Z we get the elliptic surface (elliptic curve over Q(T )) T Z(T + Z)(T + Z) = V. Now substituting 4T Z = X we obtain the elliptic surface in the Weierstrass form (15) X + 6T X + 8T 6 X = Y. The rank of curve (15) over Q(T ) is equal to 0. But standard conjectures predict that for infinitely many specializations T = t the specialized curve has positive rank over Q. We list some small solutions (ε φ γ) of equation (14): ( 180 900 196000) ( 160 1600 456000) ( 15 89 584) ( 16 89 650) ( 15 5 60750) ( 108 144 7776) ( 96 144 184) ( 90 100 750) ( 90 5 9115) ( 48 144 7648) ( 6 144 8) ( 18 9 81) ( 10 100 6750) ( 6 9 7) ( 9 54) (7 144 46656) (00 1600 40000). Finally let us find the example for the case A4). For a e n u Q[X] it must hold ae + n = u. Thus let us find an identity of the form (16) (X 4 +α X +α X +α 1 X+α 0 )(ϑ µ )+(µ X +µ 0 ) = (ϑ X +ϑ 1 X+ϑ 0 ).

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS 9 From (16) we obtain α = ϑ ϑ 1 ϑ α = ϑ 1 µ µ 0 + ϑ ϑ 0 µ ϑ α 1 = ϑ 1ϑ 0 µ ϑ µ and α 0 = ϑ 0 µ 0 ϑ. Similarly from be + n = v where b = X 4 + β X + β X + µ β 1 X + β 0 and v = ν X + ν 1 X + ν 0 are polynomials over Q we see that ν = ϑ and then we get β = ϑ ν 1 ϑ β = ν 1 µ µ 0 + ϑ ν 0 µ ϑ β 1 = ν 1ν 0 µ ϑ µ β 0 = ν 0 µ 0 ϑ. From the condition that (X 4 +α X +α X +α 1 X+α 0 )(X 4 +β X + µ β X + β 1 X + β 0 ) + µ X + µ 0 is a square of a polynomial of degree 4 we have a system of three equations. For ϑ 1 = ν 1 = µ 0 = 0 one of the equations has a rational solution if ϑ 0 ν 0 is a square. Thus we assume that ν 0 = ϑ 0 τ where τ Q and then ϑ = µ (1 + τ ). It remains to solve the equation τ 9 ϑ τ 0+µ 4 τ 8 +64τ 7 ϑ 0 4µ 4 τ 6 + τ 5 ϑ 0 + 6µ 4 τ 4 4µ 4 τ + µ 4 = 0 which has a solution if ϑ 0τ(1 + τ ) is a fourth power. Therefore we take ϑ 0 = τ(1 + τ ). We obtain µ = 4τ τ 4 + τ + 1 (τ 1)(τ + 1) and then a = X 4 1 + τ τ 1 X + 1 b = X 4 τ 1 + τ τ 1 X + τ 4 e = 4τ (τ 4 + τ + 1) n = 4τ τ 4 + τ + 1 (τ 1)(τ + 1) X. For τ = we obtain polynomials over Q and multiplying them by 1 we get that {4(X 1)(X ) 4(X 1)(X 4) 5(X 5)(X )} is a polynomial D(1900X )-triple with e = 1780000. Extending this triple to the regular quadruple in Q[X] we get the fourth element d = 1 1 X (X )(X + )(X 1)(X 19)(X 7).. Diophantine quadruples for products of twin primes By considering congruences modulo 4 it is easy to prove that if n is an integer of the form n = 4k + then there does not exist a Diophantine quadruple with the property D(n) [ 1 5]. On the other hand it was proved in [5] that if an integer n does not have the form 4k+ and n S 1 = { 4 1 5 8 1 0} then there exists at least one Diophantine quadruple with the property D(n). Moreover if n (mod 4) and n S where S = S 1 { 15 1 7 7 1 15 1 4 8 48 60 84} then there exist at least two distinct D(n)-quadruples. Thus it seems natural to ask whether similar results hold for integers n with at least l distinct D(n)-quadruples where l. An experimental search for quadruples with elements of small size together with some theoretical observations from [8 9] suggest that the answer is negative. Conjecture. The set S of all integers n not of the form 4k + with the property that there exist at most two different D(n)-quadruples is infinite. By the results from [8] we know that if an integer n is not of the form 4k + n > 48 and there exist at most two distinct D(n)-quadruples then n has one of the following forms: 4k + 16k + 1 8k + 5 k + 0. Since multiplying all elements of D(4k + ) and D(8k + 5)-quadruples by we obtain D(16k + 1) and D(k + 0)-quadruples respectively (by [5 Remark ] all D(16k + 1)-quadruples

10 ANDREJ DUJELLA AND ANA JURASIĆ can be obtained on this way) we may restrict our attention to the integers of the forms 4k + and 8k + 5. We now quote main result of [9]. Lemma 1. (i) Let n be an integer such that n (mod 4) n { 9 1 7 11} and there exist at most two distinct D(n)-quadruples. Then n 1 n 9 and 9n 1 are primes. Furthermore either n is prime or n = pq where p and q = p + are twin primes. (ii) Let n be an integer such that n 5 (mod 8) n { 7 5 1 1 45} and there exist at most two distinct D(n)-quadruples. Then the integers n n 1 4 n 9 4 and 9n 1 4 are primes. Note that by Dickson s conjecture on simultaneous prime values of linear polynomials and its extension by Schinzel and Sierpiński [6] it is predicted that there are infinitely many integers n satisfying the primality conditions from Lemma 1. Note also that there is one notable difference between (i) and (ii) parts of Lemma 1 which concerns twin primes occurring in (i). However the preliminary version of (ii) also contained similar condition concerning the numbers n of the form n = pq where p and q are primes such that q = p + 4. But it was possible to remove this case because of the following polynomial formula: the set { X + X + 10 (1X + 11)(1X + 17X + 7X + 1) (1X + 1)(1X + 19X + 9X + 1)} has the property D((6X + 1)(6X + 5)). Thus we may ask whether the condition concerning the product of twin primes can be removed from the statement (i) of Lemma 1. We will give partial positive answer to this question. Namely we will show that the condition can be removed if p 1 (mod 4). Let us recall that in [5] it was proved that for an integer k { 1 0 1 } the sets {1 k k k +1 4k 4k } and {1 9k +8k+1 9k +14k+6 6k +44k+1} are two distinct D(4k + )-quadruples. If p and q are twin primes then p = k 1 q = k + 1 for a positive integer k and n = pq = 4k 1. Thus our goal is to find a new polynomial quadruple with property D(4X 1). Motivated by the following example {1 1667501 1846 699955} of a D(41 4)-quadruple (note that for n = 41 4 = 176 the numbers n 1 = 881 n 9 = 877 and 9n 1 = 79 are primes) we search for a polynomial quadruple of the form {1 a b a + b + r} where a and b are polynomials of degree 4. In search for more general result we take n = αx + β. Since the polynomial n is even we assume that a( X) = b(x). Let us put (17) (18) a + n = (α X α 1 X + α 0 ) b + n = (α X + α 1 X + α 0 ). We express a and b from (17) and (18). Then ab + n is a polynomial of degree 8 which should be a perfect square. Thus we put (19) ab + n = (α X 4 + ( α + α 0 α α 1)X + γ). This gives us three equations in α 0 α 1 α α β γ. Let r = α X 4 + ( α + α 0 α α 1)X + γ. Consider the set {1 a b a + b + r}. In order to satisfy the definition

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS 11 of a D(n)-quadruple it remains to satisfy the condition that a + b + r + n is a perfect square. Namely by (17) and (18) 1 a + n and 1 b + n are squares and we have ab + n = r a(a + b + r) + n = (a + r) b(a + b + r) + n = (b + r). From ( a+b+r+n = 4αX 4 +( α+8α 0 α )X β+γ+α0 = α X + α + 8α ) 0α 4α we get γ = 16α β + αα 0 + 9α 48α α 0 α α. Inserting this in the three equations induced by (19) we get α 0 = 9α + 78α1α + 64α1 4 6α1 α α = 4α + 16α1 9 β = (7α + 198α1α + 40α1α 4 + 00α1) 6 81α1. α In that way we obtained two parametric families of a polynomial quadruples of the desired type. It remains to choose the parameters α 1 and α in order to obtain D(4X 1)-quadruples. We get the following pairs (α 1 α ) = (± 15 ) (± ) (± 1 5 ). The last task is to get integer elements of the quadruple. By taking (α 1 α ) = ( ) we are almost done since we get a = 9X4 9X X + 5 4 X + 5 4 b = 9X 4 + 9X + X + 5 4 X + 5 4 a + b + r = 6X4 4X + 1 i.e. the set {1 9X 4 9X X + 5 4 X + 5 4 9X4 + 9X + X + 5 4 X + 5 4 6X4 4X + 1} is the Diophantine quadruple with the property D(4X 1). For X odd say X = k + 1 numbers a and b are integers. Therefore we get that if k { 1 0} is an integer then (0) {1 144k 4 + 16k + 11k + 0k + 1 144k 4 + 60k + 9k + 14k + 576k 4 + 115k + 848k + 7k + } is a D((4k + 1)(4k + ))-quadruple. Thus we proved the following extension of Lemma 1 (i). Theorem 1. Let n be an integer such that n (mod 4) n { 9 1 7 11} and there exist at most two distinct D(n)-quadruples. Then n 1 n 9 and 9n 1 are primes. Furthermore either n is prime or n = pq where p (mod 4) and q = p + are twin primes. 4. Adjugates of D(n)-quadruples In [0] Gibbs described the following construction by which from one Diophantine quadruple we can obtain up to eight new Diophantine quadruples. Given a (possibly improper) Diophantine quadruple {a 1 a a a 4 } with the property D(n) such that a i a j + n = x ij an adjugate quadruple is obtained by constructing the symmetric matrix S with components S ii = a i and S ij = x ij for i j and forming its adjugate matrix T. The adjugate quadruple {A 1 A A A 4 } is a (possibly improper) Diophantine quadruple such that A i A j + N = Xij where A i = T ii X ij = T ij and N = n det(s). A quadruple has at most eight distinct adjugates

1 ANDREJ DUJELLA AND ANA JURASIĆ which arise from different combinations of the signs of x ij. Gibbs also stated the following relationships between properties of a quadruple and its adjugates: A) If a quadruple is regular with property D(1) then four of its eight adjugates are the regular quadruples formed by dropping one of its four components and replacing it with the other extension value. B) If a quadruple is semi-regular then four of its eight adjugates will be improper quadruples with zero component. C) If a quadruple is twice semi-regular then two of its adjugates have two zeros and four others have one zero component. D) If a quadruple is improper with a zero component then all of its adjugates include a regular triple. E) If a quadruple is improper with two duplicate components then it can have at most six distinct adjugates and four of them are improper with two duplicate components. F) If a quadruple is improper with two zero components then all of its adjugates are twice semi-regular but four of them are improper with two duplicate components and there are only two other distinct adjugates. Gibbs gave numerical examples for each of the cases A) F). We will give polynomial examples for all cases. In some of them we will use quadruples constructed in the previous sections. We will give details on quadruples obtained by Gibbs construction only in the cases where this construction produces Diophantine quadruples for quadratic polynomials. A) The set {X 1 X + 1 4X(X 1)(X + 1) 8X(X 1)(X + 1)(X 1)} is a regular polynomial D(1)-quadruple (see []). By the above construction we obtain: four regular D(1)-quadruples and irregular quadruples with the properties D(56X 6 56X 5 18X 4 + 18X + 16X 8X + 1) D(56X 6 + 56X 5 18X 4 18X + 16X + 8X + 1) D(4096X 8 510X 6 + 190X 4 4X + 9) and D( 1684X 10 + 867X 8 16896X 6 + 968X 4 0X + 9). B) By applying Gibbs construction on the quadruple (0) which is a semiregular polynomial D((4X + 1)(4X + ))-quadruple {1 (1X + 1)(1X + 17X + 8X + 1) (1X + 11)(1X + 19X + 10X + ) we obtain: (4X + 16X + )(4X + X + 11)} four improper irregular quadruples and (proper) irregular quadruples with the properties D((4X + 1) (4X + ) (8944X 8 + 1776X 7 + 560448X 6 + 5018X 5 + 8880X 4 + 9675X + 1898X + 1944X + 81)) D((X + 1) (4X + 1) (4X + )(1776X 7 + 10787X 6 + 14584X 5 + 1047744X 4 + 4058X + 98800X + 1156X + 507)) D((X + 1) (4X + 1)(4X + ) (1776X 7 + 144160X 6 + 1951488X 5 + 1659456X 4 + 8451X + 7776X + 684X + 161)) and

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS 1 D( (4X + 1) (4X + ) (119496X 10 + 59719680X 9 + 1795X 8 + 17517778X 7 + 14976016X 6 + 86878656X 5 + 4567880X 4 + 97656X + 15900X + 15664X + 6471)). C) The set {4X 16X 6X 1 4X 4} obtained from (1) is a twice semi-regular polynomial D(4X + 1)-quadruple from which we obtain: six improper irregular quadruples and irregular quadruples with the properties D((4X + 1)(6144X 6656X + 616X 79)) and D( (4X + 1) (6144X 5184X + 816X 1)). D) In A) we obtained the set {X+1 4(X+) (X+5) 0} which is an improper regular polynomial D(1)-quadruple. For this set using Gibbs construction we obtain: four improper regular quadruples a regular D(1)-quadruple and the following (proper) quadruples for quadratic polynomials: {X +1 8(X +) X + 4(X +)(X +4)(6X +11)} is a semi-regular D(48X + 16X + 97)-quadruple {5(5X + 9) 4(X + ) X + 4(X + )(X + 4)(6X + 11)} is a semiregular polynomial D(96X + 0X + 65)-quadruple {5(5X + 9) 8(X + ) (X + 5) 4(X + )(X + 4)(6X + 11)} is a semi-regular polynomial D( (48X + 15X + 117))-quadruple. E) The set { X X +4 X +4} is an improper irregular polynomial D(X +9)- quadruple (see [5]) which includes three regular triples. By the above construction we obtain six improper irregular quadruples. However if we take the set {(X 1)(X + 1)(5X + ) (X 1)(X + 1)(5X + ) (X 1)(X + 1)(5X + ) X (7X + 1)} which is an improper irregular polynomial D(4X (5X + ) 4 )-quadruple (obtained using [6 Proposition ]) the above construction gives: four improper irregular quadruples and irregular quadruples with the properties D(4X (5X + ) (5X 6 + 70X 4 X + 4) and D(4X 4 (5X + ) (14X 6 6X 4 6X + 9)). Let us mention that it is possible to find a polynomial quadruple with two pairs of equal elements. E.g. {4X 4X (X X 1)(X + X 1) (X X 1)(X + X 1)} is an improper twice semi-regular polynomial D(4(X 1) X (X + 1) )-quadruple (see [17]). Here Gibbs construction gives five improper irregular quadruples. F) The set { 4(X +5) (X +6) 0 0} obtained in E) is an improper irregular polynomial D((X + 9) )-quadruple. For this set we obtain: four improper twice semi-regular quadruples and the following (proper) quadruples for quadratic polynomials: { (7X + 6) 5(X + 16) (X + 4)} is a twice semi-regular polynomial D((X + 9)(8X + 41))-quadruple

14 ANDREJ DUJELLA AND ANA JURASIĆ {X (4X + 19) 5(X + 16) X + 4} is a twice semi-regular polynomial D((X + 9)(1X + 49))-quadruple. Acknowledgement: The authors were supported by the Ministry of Science Education and Sports Republic of Croatia grant 07-07781-81. References [1] A. Baker and H. Davenport The equations x = y and 8x 7 = z Quart. J. Math. Oxford Ser. () 0 (1969) 19-17. [] E. Brown Sets in which xy + k is always a square Math. Comp. 45 (1985) 61-60. [] Y. Bugeaud A. Dujella and M. Mignotte On the family of Diophantine triples {k 1 k + 1 16k 4k} Glasgow Math. J. 49 (007) -44. [4] Diophantus of Alexandria Arithmetics and the Book of Polygonal Numbers (I. G. Bashmakova Ed.) (Nauka 1974) 10-104. [5] A. Dujella Generalization of a problem of Diophantus Acta Arith. 65 (199) 15-7. [6] A. Dujella On Diophantine quintuples Acta Arith. 81 (1997) 69-79. [7] A. Dujella The problem of the extension of a parametric family of Diophantine triples Publ. Math. Debrecen 51 (1997) 11-. [8] A. Dujella Some estimates of the number of Diophantine quadruples Publ. Math. Debrecen 5 (1998) 177-189. [9] A. Dujella A problem of Diophantus and Dickson s conjecture Number Theory Diophantine Computational and Algebraic Aspects (K. Gyory A. Petho V. T. Sos eds.) Walter de Gruyter Berlin 1998 147-156. [10] A. Dujella There are only finitely many Diophantine quintuples J. Reine Angew. Math. 566 (004) 18-14. [11] A. Dujella and C. Fuchs Complete solution of the polynomial version of a problem of Diophantus J. Number Theory 106 (004) 6-44. [1] A. Dujella C. Fuchs and F. Luca A polynomial variant of a problem of Diophantus for pure powers Int. J. Number Theory 4 (008) 57-71. [1] A. Dujella C. Fuchs and R. F. Tichy Diophantine m-tuples for linear polynomials Period. Math. Hungar. 45 (00) 1-. [14] A. Dujella C. Fuchs and G. Walsh Diophantine m-tuples for linear polynomials. II. Equal degrees J. Number Theory 10 (006) 1-8. [15] A. Dujella and A. Jurasić On the size of sets in a polynomial variant of a problem of Diophantus Int. J. Number Theory 6 (010) 1449-1471. [16] A. Dujella and A. Pethő A generalization of a theorem of Baker and Davenport Quart. J. Math. Oxford Ser. () 49 (1998) 91-06. [17] A. Dujella and V. Petričević Strong Diophantine triples Experiment. Math. 17 (008) 8-89. [18] Y. Fujita The extensibility of Diophantine pairs {k 1 k+1} J. Number Theory 18 (008) -5. [19] Y. Fujita Any Diophantine quintuple contains a regular Diophantine quadruple J. Number Theory 19 (009) 1678-1697. [0] P. Gibbs Adjugates of Diophantine Quadruples Integers 10 (010) 01-09. [1] H. Gupta and K. Singh On k-triad sequences Internat. J. Math. Math. Sci. 5 (1985) 799-804. [] B. W. Jones A variation of a problem of Davenport and Diophantus Quart. J. Math. Oxford Ser.() 7 (1976) 49-5. [] B. W. Jones A second variation of a problem of Davenport and Diophantus Fibonacci Quart. 15 (1977) -0. [4] A. Jurasić Diophantine m-tuples for quadratic polynomials Glas. Mat. Ser. III to appear. [5] S. P. Mohanty and A. M. S. Ramasamy The characteristic number of two simultaneous Pell s equations and its application Simon Stevin 59 (1985) 0-14. [6] A. Schinzel and W. Sierpiński Sur certaines hypotheses concernant les nombres premiers Acta Arith. 4 (1958) 185-08 Corrigendum 5 (1959) 59.

SOME DIOPHANTINE TRIPLES AND QUADRUPLES FOR QUADRATIC POLYNOMIALS 15 A. Dujella Department of Mathematics University of Zagreb Bijenička cesta 0 10000 Zagreb Croatia duje@math.hr A. Jurasić Department of Mathematics University of Rijeka Omladinska 14 51000 Rijeka Croatia ajurasic@math.uniri.hr