ON THE EXCEPTIONAL SET IN THE PROBLEM OF DIOPHANTUS AND DAVENPORT

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1 ON THE EXCEPTIONAL SET IN THE PROBLEM OF DIOPHANTUS AND DAVENPORT Andrej Dujell Deprtment of Mthemtics, University of Zgreb, Zgreb, CROATIA The Greek mthemticin Diophntus of Alexndri noted tht the numbers x, x + 2, 4x + 4 nd 9x + 6, where x = 1, hve the following property: the 16 product of ny two of them incresed by 1 is squre of rtionl number (see [4]). Fermt first found set of four positive integers with the bove property, nd it ws {1, 3, 8, 120}. Lter, Dvenport nd Bker [3] showed tht if d is positive integer such tht the set {1, 3, 8, d} hs the property of Diophntus, then d hs to be 120. In [2] nd [5], the more generl problem ws considered. Let n be n integer. A set of positive integers { 1, 2,..., m } is sid to hve the property D(n) if for ll i, j {1, 2,..., m}, i j, the following holds: i j + n = b 2 ij, where b ij is n integer. Such set is clled Diophntine m-tuple. If n is n integer of the form 4k + 2, k Z, then there does not exist Diophntine qudruple with the property D(n) (see [2, Theorem 1], [5, Theorem 4] or [8, p. 802]). If n integer n is not of the form 4k + 2 nd n S = { 4, 3, 1, 3, 5, 8, 12, 20}, then there exists t lest one Diophntine qudruple with the property D(n), nd if n S T, where T = { 15, 12, 7, 7, 13, 15, 21, 24, 28, 32, 48, 60, 84}, then there exist t lest two distinct Diophntine qudruples with the property D(n) (see [5, Theorems 5 nd 6] nd [6, p. 315]). For n S the question of the existence of Diophntine qudruples with the property D(n) is still unnswered. This question is t present fr from being solved. Remrk 3 from [5] reduces the problem to the elements of the set S = { 3, 1, 3, 5, 8, 20}. Let us mention tht in [2] nd [11], it ws proved tht the Diophntine triples {1, 2, 5} nd {1, 5, 10} with the property D( 1) cnnot be extended to the Diophntine qudruples with the sme property. Our hypothesis is tht for n S there does not exist Diophntine qudruple with the property D(n). In this pper we consider some consequences of this hypothesis to the problem of Diophntus for liner polynomils. 1

2 PROBLEM OF DIOPHANTUS AND DAVENPORT 2 Definition 1 Let k 0 nd l be integers. A set of liner polynomils with integrl coefficients { i x+b i : i = 1, 2,..., m} is clled liner Diophntine m-tuple with the property D(kx + l) if ( i x + b i )( j x + b j ) + kx + l is squre of polynomil with integrl coefficients for ll i, j {1, 2,..., m}, i j. We cll liner Diophntine m-tuple cnonicl if gcd( 1, 2,..., m, k) = 1. Remrk 1 If the set { i x + b i : i = 1,..., m} is liner Diophntine m-tuple with the property D(kx + l), then the numbers 1,..., m re ll of the sme sign. Therefore we my ssume tht 1,..., m re positive. If the bove m-tuple is cnonicl, then the numbers 1,..., m re perfect squres. If gcd( 1,..., m, k) = e > 1, then replcing ex by x we get cnonicl liner Diophntine qudruple with the property D( kx + l). e Some spects of the problem of Diophntus for polynomils were considered in [1], [5], [7], [9] nd [10]. In [5], it ws proved tht if { i x + b i : i = 1, 2, 3, 4} is liner Diophntine qudruple with the property D(kx + l), then k is even, nd if the bove qudruple is cnonicl nd if gcd(k, l) = 1, then l is qudrtic residue modulo k. It is not known whether the converse of this result is true. We will show tht this question is connected with the our hypothesis bout the elements of the exceptionl set S. The bsic ide is to consider liner Diophntine qudruples which hve two elements with equl constnt terms. Lemm 1 Let { 2 x, b 2 x } be liner Diophntine pir with the property D(kx + l). Then there exists n integer α such tht l = α 2 2 nd k = ( 2 + b 2 ) + 2αb. Proof: Since 2 + l is perfect squre, we conclude tht there exists n integer α such tht l = α 2 2. From it follows tht k = ( 2 + b 2 ) + 2αb. ( 2 x )(b 2 x ) + kx + l = (bx + α) 2 Lemm 2 Let { 2 x, b 2 x, c(x)} be liner Diophntine triple with the property D(kx + l). In the nottion of Lemm 1, we hve: c(x) {( + b) 2 x + 2α 2, ( b) 2 x 2α 2, k 2k(α ) [ ( + b) ]2 x + ( + b), [ k 2k(α + ) 2 ( b) ]2 x ( b) }. 2

3 PROBLEM OF DIOPHANTUS AND DAVENPORT 3 Proof: Write c(x) = c 2 x γ. Then there exists n integer δ such tht γ + α 2 2 = δ 2. (1) We conclude from ( 2 x )(c 2 x γ) + kx + l = (cx ± δ) 2 nd Lemm 1 tht Combining (2) with (1) we obtin nd finlly γ 2 c 2 + ( 2 + b 2 ) + 2αb = ±2δc. (2) (α + b) 2 = (δ ± c) 2 α + b = ±δ ± c. (3) In the sme mnner we cn see tht from (b 2 x )(c 2 x γ) + kx + l = (bcx ± δ) 2 it follows tht αb + = ±δb ± c. (4) Solving the systems of the equtions (3) nd (4) we get ( c, δ ) {( + b, α ), ( b, α + ), k )( b) k + )( + b) (, (α ), (, (α )}. ( + b) + b ( b) b From (1) we see tht c(x) {( + b) 2 x + 2α 2, ( b) 2 x 2α 2, k 2k(α ) [ ( + b) ]2 x + ( + b), [ k 2k(α + ) 2 ( b) ]2 x ( b) }. 2 Lemm 3 Let { 2 x, b 2 x, c(x), d(x)} be liner Diophntine qudruple with the property D(kx + l), where gcd(k, l) = 1. In the nottion of Lemm 1, we hve: ± 2α ± 2α,, 2α ± 3 2α ± 3,, 3 ± 2α ± 2α, }. 3

4 PROBLEM OF DIOPHANTUS AND DAVENPORT 4 Proof: Set p 1 (x) = ( + b) 2 x + 2α 2, p 2 (x) = ( b) 2 x 2α 2, p 3 (x) = k [ (+b) ]2 x + 2k(α ) k, p (+b) 2 4 (x) = [ ( b) ]2 x 2k(α+). According to Lemm 2, we hve ( b) 2 {c(x), d(x)} {p 1 (x), p 2 (x), p 3 (x), p 4 (x)}. Thus we need to consider six cses. We cn ssume tht gcd(, b) = 1, since otherwise we put x = e 2 x, where e = gcd(, b). Cse 1. {c(x), d(x)} = {p 1 (x), p 2 (x)} If y is n integer such tht (2α 2)( 2α 2) + l = y 2, then y 2 = 3l. (5) From p 1 (x) p 2 (x) + kx + l = [( 2 b 2 )x + y] 2 it follows tht (2α 2)( b) 2 (2α + 2)( + b) 2 + ( 2 + b 2 ) + 2αb = 2y( 2 b 2 ). This gives 3k = 2y( 2 b 2 ). (6) Therefore y = 3, by (5), (6) nd gcd(k, l) = 1. We conclude tht l = 3 nd tht α = 1, = 2. Combining k = ±2( 2 + b 2 ) ± 2b with (6) we get b {± 1 2, ±2}. It is esily seen tht in ll of these four cses the intersection {c(x), d(x)} { 2 x, b 2 x } is nonempty, which contrdicts our ssumption tht { 2 x, b 2 x, c(x), d(x)} is qudruple. Therefore the first cse is impossible. Cse 2. We hve: {c(x), d(x)} = {p 1 (x), p 3 (x)} 2α, 2α, 2α 3 2α 3, }. Cse 3. We hve: Cse 4. We hve: {c(x), d(x)} = {p 2 (x), p 4 (x)} 2α 2α,, 2α + 3 {c(x), d(x)} = {p 1 (x), p 4 (x)} 2α, 3 2α 2α + 3, }. 2α,, 2α 3 }.

5 PROBLEM OF DIOPHANTUS AND DAVENPORT 5 Cse 5. We hve: Cse 6. We hve: {c(x), d(x)} = {p 2 (x), p 3 (x)} 2α, 3 2α, 2α, b {c(x), d(x)} = {p 3 (x), p 4 (x)} 2α {, 2α, 2α, 2α }. 3 2α }. We give the proof only for the cse 6, which is the most involved; the proofs of the other cses re similr in spirit. Let y be n integer such tht 4k 2 ( 2 α 2 ) 2 ( 2 b 2 ) 2 + l = y 2 2 ( 2 b 2 ) 2. (7) From p 3 (x) p 4 (x) + kx + l = [ k 2 2 ( 2 b 2 ) x + y ( 2 b 2 ) ]2 it follows tht Combining (8) with (7) we hve 3 ( 2 b 2 ) 2 4k 2 = 2ky. (8) [4k 2 2 ( 2 b 2 ) 2 ] [4k 2 α 2 4 ( 2 b 2 ) 2 ] = 0. Let 4k 2 = 2 ( 2 b 2 ) 2. We cn ssume tht 2k = ( 2 b 2 ). We conclude tht 2( 2 + b 2 ) + 4αb = ( 2 b 2 ), nd hence tht 2 + 4αb + 3b 2 = 0. (9) From this we hve b = 2α±z, where z 2 = 4α Write 1 b 1 = 2α+z, 2 b 2 = 2α z, where gcd( 1, b 1 ) = gcd( 2, b 2 ) = 1, nd 2k i = ( 2 i b 2 i ), i = 1, 2. We clim tht gcd(k i, l) > 1 for i = 1, 2. Suppose, contrry to our clim, tht gcd(k i, l) = 1 for some i {1, 2}. We hve 4k 1 k 2 = 2 ( 2 1 b 2 1)( 2 2 b 2 2) = 2 b2 1b [( 2α + z)2 2 ] [( 2α z) 2 2 ] = b2 1b [(2α + )2 z 2 ] [(2α ) 2 z 2 ] = b2 1b ( 2 α 2 ) = 16lb 2 1b 2 2.

6 PROBLEM OF DIOPHANTUS AND DAVENPORT 6 We conclude from ( 2α+z)( 2α z) = 3 2 tht b 1 b 2, nd hence tht k 1 k l. Set c i = 2 i b 2 i. Since gcd(k i, l) = 1 nd the integer 2k i = c i divides 8 2, we hve 8 0 (mod c i ). From (9) it follows tht 8α i b i = 8b i 2c i 0 (mod c i ). We conclude from gcd( i, b i ) = 1 tht gcd( i, b i, c i ) = 1, nd hence tht 8α 0 (mod c i ). Thus we hve 2k 0 (mod c i ), 8l 0 (mod c i ) nd gcd(k, l) = 1, which implies c i 8, i.e. c i {±1, ±2, ±4, ±8}. Since i 0 nd b i 0, it follows tht c i = ±8. Hence α nd re odd nd l is even, which contrdicts the fct tht k is even nd gcd(k, l) = 1. Let 4k 2 α 2 = 4 ( 2 b 2 ) 2. We hve [2kα + 2 ( 2 b 2 )] [2kα 2 ( 2 b 2 )] = [ + (2α + )b] [(2α ) + b] [ + (2α )b] [(2α + ) + b] = 0. Hence b 2α {, 2α, 2α, 2α }. Theorem 1 Let l { 3, 1, 3, 5}. Write e( 3) = 21, e( 1) = 5, e(3) = 39 nd e(5) = 55. Suppose tht there does not exist Diophntine qudruple with the property D(l). Then there does not exist Diophntine qudruple with the property D(kx + l), provided gcd(k, e(l)) = 1. Proof: Let l { 3, 1, 3, 5} nd let k be n integer such tht gcd(k, l) = 1. Suppose tht { i x + b i : i = 1, 2, 3, 4} is cnonicl liner Diophntine qudruple with the property D(kx + l). Then the set {b 1, b 2, b 3, b 4 } hs the property tht the number b i b j + l is perfect squre for ll i, j {1, 2, 3, 4}, i j. Since, by ssumption of the theorem, this set is not Diophntine qudruple with the property D(l), we conclude tht there exist indices i, j {1, 2, 3, 4}, i j, such tht b i = b j. Without loss of generlity we cn ssume tht b 1 = b 2 =. The integers l hve the unique representtion s difference of the squres of two integers: 3 = , 1 = , 3 = , 5 = From Lemm 3 by n esy computtion we conclude tht for l { 3, 1, 3, 5} there is one nd only one cnonicl liner Diophntine qudruple with the property D(kx + l), such tht gcd(k, l) = 1. These qudruples re {4x 2, 9x 2, 25x 6, 49x 14}, (10)

7 PROBLEM OF DIOPHANTUS AND DAVENPORT 7 {x 1, 9x 1, 16x 2, 25x 5}, (11) {9x + 1, 25x + 1, 64x + 6, 169x + 13}, (12) {9x + 2, 16x + 2, 49x + 10, 121x + 22} (13) with the properties D(14x 3), D(10x 1), D(26x+3) nd D(22x+5) respectively. This proves the theorem. Remrk 2 The sets (10) (13) re the specil cses of the following more generl formul from [7]: the set {9m + 4(3k 1), (3k 2) 2 m + 2(k 1)(6k 2 4k + 1), (3k + 1) 2 m + 2k(6k 2 + 2k 1), (6k 1) 2 m + 4k(2k 1)(6k 1)} (14) hs the property D(2m(6k 1) + (4k 1) 2 ). The sets (10) (13) cn be obtined from (14) for k = 1, m = x + 2; k = 1, m = x 1; k = 2, m = x + 3 nd k = 2, m = x 2 respectively. References [1] J. Arkin & G. E. Bergum. More on the problem of Diophntus. In Appliction of Fiboncci Numbers 2: Ed. A. N. Philippou, A. F. Hordm & G. E. Bergum. Dordrecht: Kluwer, [2] E. Brown. Sets in which xy + k is lwys squre. Mthemtics of Computtion 45 (1985): [3] H. Dvenport & A. Bker. The equtions 3x 2 2 = y 2 nd 8x 2 7 = z 2. Qurt. J. Mth. Oxford Ser. (2) 20 (1969): [4] Diofnt Aleksndriĭskiĭ. Arifmetik i knig o mnogougol nyh chislkh. Moscow: Nuk, [5] A. Dujell. Generliztion of problem of Diophntus. Act Arithmetic 65 (1993): [6] A. Dujell. Diophntine qudruples for squres of Fiboncci nd Lucs numbers. Portuglie Mthemtic 52 (1995):

8 PROBLEM OF DIOPHANTUS AND DAVENPORT 8 [7] A. Dujell. Some polynomil formuls for Diophntine qudruples. Grzer Mthemtishe Berichte (to pper). [8] H. Gupt & K. Singh. On k-trid sequences. Internt. J. Mth. Mth. Sci. 8 (1985): [9] B. W. Jones. A vrition on problem of Dvenport nd Diophntus. Qurt. J. Mth. Oxford Ser. (2) 27 (1976): [10] B. W. Jones. A second vrition on problem of Diophntus nd Dvenport. The Fiboncci Qurterly 16 (1978): [11] S. P. Mohnty & M. S. Rmsmy. The simultneous Diophntine equtions 5y 2 20 = x 2 nd 2y 2 +1 = z 2. Journl of Number Theory 18 (1984): AMS Clssifiction Numbers: 11D09, 11C08