ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II ALAIN TOGBÉ

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1 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II ALAIN TOGBÉ 1

2 2 ALAIN TOGBÉ On the solutions of a family of quartic Thue equations II Alain Togbé Purdue University North Central, Westville Indiana April 12, 2006 Number Theory Seminar Purdue University, West Lafayette Indiana

3 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 3 1. Diophatine equations Definitions (1) A diophantine equation is an equation of the form P (x 1,..., x d ) = 0, d N (2) If P (a 1,..., a d ) = 0 where a i Z for i = 1,..., d, then (a 1,..., a d ) is a solution. (3) A Thue equation is an equation of the form F (x, y) = a d x d +a d 1 x d 1 y + +a 1 xy d 1 +a 0 y d = A, where a i, A Z with A 0.

4 4 ALAIN TOGBÉ The history of Diophantine equations is very rich and goes back to the antiquity. But Thue was first to study an infinite parametrized family of Thue equations. He proved that (a + 1)X n ay n = 1, x > 0, Y > 0 has only the solution x = y = 1 for a suitably large in relation to prime n 3. There are many methods to solve families of Thue equations: (1) Baker s method, (2) The hypergeometric method.

5 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 5 2. Baker s method Let us consider a family of parameterized Thue equations F (x, y) = A (with a parameter n). (1) Determine the obvious solutions of the equation. (2) Define the number field K = Q(θ), where θ is a root of the polynomial F (x, 1). (3) Determine a system of independent units {ε 1,..., ε r } of K. Therefore, there exist integers u 1,..., u r such that (x θy) I = ε u εu r r where r is the rank of K. (4) Estimate the regulator R of K. (5) Find an upper bound for U = max{u 1,..., u r }, in terms of R and log y.

6 6 ALAIN TOGBÉ (6) Compute an upper bound of the linear form Λ in logarithms obtained by a Siegel s formula. (7) Combine the results of steps 4, 5, 6 to find an upper bound for Λ in terms of U. (8) Determine a lower bound for U. (9) Combine the results of steps 7 and 8 to obtain a negative upper bound for log Λ. (10) Determine a negative lower bound for log Λ. (11) Use the lower and upper bounds obtained from 9 and 10 for log Λ to compute a numerical upper bound for the parameter n, say n N.

7 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 7 3. A survey A list of families of Thue equations studied since 1990 can be obtained at: cheub/thue.html Families of degree 3 The first family of Thue equations is solved by Emery Thomas in 1990: x 3 (a 1)x 2 y (a + 2)xy 2 y 3 = 1 The solutions are: (0, 1), (1, 0), ( 1, 1). Other cubic families of Thue equations are solved by Thomas, Mignotte-Tzanakis, Wakabayashi, and T. x 3 (n 3 2n 2 + 3n 31)x 2 y n 2 xy 2 y 3 = ±1, The solutions are: (0, ±1), (±1, 0).

8 8 ALAIN TOGBÉ Families of degree 4 The first quartic family of Thue equations is solved by Attila Pethő in 1991: x 4 ax 3 y x 2 y 2 + axy 3 + y 4 = ±1; the solutions are: ±{(0, 1), (1, 0), (1, 1), (1, 1), (a, 1), (1, a)}. Many quartic families are solved by: Pethő, Mignotte- Pethő-Roth, Lettl-Pethő, Chen-Voutier, Wakabayashi, Heuberger-Pethő-Tichy, Dujella-Jadrijevič, and T. x 4 n 2 x 3 y (n 3 + 2n 2 + 4n + 2)x 2 y 2 n 2 xy 3 + y 4 = 1; for n and n with n, n + 2, n squarefree. The solutions are: ±{(0, 1), (1, 0)}.

9 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 9 Families of degree 5 The first quintic family of Thue equations is solved by Clemens Heuberger in 1998: x(x 2 y 2 )(x 2 a 2 y 2 ) y 5 = ±1, for a > The solutions are {(0, ±1), (±1, 0), (1, ±1), ( 1, ±1), (a, ±1), ( a, ±1)}. Two other families are solved by Gaál-Lettl, Levesque- Mignotte.

10 10 ALAIN TOGBÉ Family of degree 6 Only one sextic family of Thue equations is solved by Lettl-Pethő-Voutier in 1999: x 6 2ax 5 y (5a + 15)x 4 y 2 20x 3 y 3 + 5ax 2 y 4 + (2a + 6)xy 5 + y 6 {±1, ±27}, with y 2 x y. The solutions are (0, 1), (1, 1). They used the hypergeometric method.

11 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 11 Family of degree 8 Only one octic family of Thue equations is solved by Heuberger-T-Ziegler in 2004: X 8 8nX 7 Y 28X 6 Y nX 5 Y X 4 Y 4 56nX 3 Y 5 28nX 2 Y 6 + 8nXY 7 + Y 8 = ±1 has only trivial solutions (±1, 0), (0, ±1) for n , where n {a Z : a + b 2 = (1 + 2) 2k+1 : k N}. Recently, Heuberger-Tethö-Tychy and Ziegler studied some families of relative Thue equations.

12 12 ALAIN TOGBÉ 4. The new result Theorem 4.1. [Togbé, 2000] For n and n with n, n+2, n 2 +4 squarefree. the equation (4.1) Φ n (x, y) = x 4 n 2 x 3 y (n 3 +2n 2 +4n+2)x 2 y 2 n 2 xy 3 +y 4 = 1. has only the trivial solutions (4.2) ±{(1, 0), (0, 1)}. For this, we used two methods: a method proposed by M. Mignotte the Bilu-Hanrot s method. Moreover, we have conjectured that (4.1) has only the solutions (4.4) for any parameter n 1.

13 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 13 Here is the new result: Theorem 4.2. [Togbé, 2006] For n 1, the family of parameterized Thue equations (4.3) Φ n (x, y) = x 4 n 2 x 3 y (n 3 +2n 2 +4n+2)x 2 y 2 n 2 xy 3 +y 4 = 1 has only the integral solutions (4.4) ±{(1, 0), (0, 1)}; except for n = 1 where we have (4.5) ±{(1, 1), (1, 0), (0, 1)}.

14 14 ALAIN TOGBÉ 5. Sketch of the proof of Theorem 4.2 First Step Lemma 5.1. Let us consider O = Z[α, β, ε], and 1, α, β, ε a subgroup of the unit group. For n 33, we have (5.1) I := [O : 1, α, β, ε ] < log 3 (n). Proof. Let R be the regulator of 1, α, β, ε : (5.2) R = 2 log(ε) ( log 2 (α) + log 2 (β) ). Using asymptotic expressions, one can check that (5.3) 10 log 3 (n) R log 3 (n), for n 33. So R > 0 and α, β, ε are independent units. We find the following bound for the index (5.4) I = [O R : 1, α, β, ε ] log3 (n). Reg Z Kn So we obtain I log 3 (n).

15 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 15 Lemma 5.2. If (x, y) is a solution of type j, then (y, x) is a solution of type j + 2. Proof. We obtain x y α i < 1. This means that x 2y 2 y is a convergent to α j. Also we have x x α j y + α j y α j y + 1 y 2 α 2 j y. Therefore we obtain (5.5) 1 x α j y = x α jy αj x 1 2 αj xy < 1 xy 1 4. This means that x y α j = min { x αi y }. 1 i 4 As (5.6) α 3 = 1, α 4 = 1 1, i.e. = α i+2 α 1 α 2 α i therefore we study the cases j = 1 and j = 4.

16 16 ALAIN TOGBÉ One can check that γ i := x α i y are units in O. Lemma 5.3. Let n 1 and (x, y) be a solution to (4.1) of type j such that y 2. Then { (5.7) γj 1 8 kj y 3, where k if j = 1, j = n 6 8 if j = 4, n 2 (5.8) log γi = log(y)+log αi α j + Proof. For i j we have y αi α j 2 γi, then (5.9) Since γj 1 = i j γi 8 y 3 α i α j i 1 i j { 1 { n 6 if j = 1, n 2 if j = 4, 2n 8 if j = 1, 1 2n 4 if j = 4. 1 αi α j. we obtain (5.7). Moreover, we know that γ i (5.10) y α i α j = 1 + γ j y(α i α j ), then taking the log of the previous expression and using expressions of α i, (5.7) we obtain (5.8).

17 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 17 Lemma 5.4. Let (x, y) be a solution to (4.1) with y 2 and n 33. Then (5.11) log y 2.485n log(n). Proof. If (x, y) is a solution to (4.1), then we determine asymptotic expressions of u 2, u 3. For each j, we define the following linear combinations of the u k such that we eliminate, if it is possible, the terms in log(n) log(y), log 2 (n), and log(n) appearing in the expressions of the Ru 2 I and Ru 3 I : (5.12) Rv j I = As y 2 and v j I Ru 2 I 2Ru 3 I = ( 4 n + 2 n 2 +L 33 ( log(2) 9 log(n) n + 2Ru 2 I + Ru 3 I = ( 4 n + 2 n 2 +L 33 ( log(n) log(5) n + )) 3n log(y)+ 3 2 log(2) 11 log(n) ( +L 0.1 ) 2n 2 33 n 2 )) 3n log(y) + 15 log 2 (n) log(n) log(5)+8 ( +L 0.1 ) 2n 2 33 n 2 is an integer, we have Rv j I R.

18 18 ALAIN TOGBÉ Lemma 5.5. For n , the equation (4.1) has no solutions, except the trivial solutions. Proof. We consider the linear forms in logarithms defined by (5.13) Λ j = Ãj log (α) + B j log (β) + log (λ j ) = log 1 + τ j. where j = 1, 4, with (5.14) Ã 1 = 2u 3 I, B1 = 2u 2 I, λ 1 = α + β α + 1/β, τ 1 = γ 1 γ 2 and (5.15) Ã 4 = 2u 2 I, B4 = 2u 3 I, λ 4 = We obtain 1/α + 1/β α + 1/β, τ 4 = γ 4 γ 3 ( β 2 ) 1, αβ + 1 (5.16) log Λj 4 log(y) + log(8) aj log(n) 2 n, with a j = 4 if j = 1 and a j = 3 if j = 4. ( ) α 1/α. α + 1/β

19 ON THE SOLUTIONS OF A FAMILY OF QUARTIC THUE EQUATIONS II 19 (5.17) log 5 Λ j (log (b ) +.14) 2 h 1 h 2 ; with (5.18) h 1 = 9 4 log(n) + 2 n, ( h 2 = b = c j3 n log(n), with for n n log(n) + 3 2n 2 log(n) 8 5n 2 log(n) c 11 = 15 4, c 12 = 49, c 13 = 0.664, c 41 = 15, c 42 = 37, c 43 = ) log(y) + c j1 log(n)+ c j2 4n,

20 20 ALAIN TOGBÉ Second Step We used a computational method based on Baker s method. For j = 1, 4 and 50 n , We developed program in Maple 9.5 and ran it on a Pentium 4 with 3.92 GHz running under Linux 7.2. First we consider j = 1, 4 (together) and 50 n , then j = 1 and n It took in average 1.75 seconds for each value of n. Third Step For 1 n 50, we use Kash to obtain the solutions in Theorem 4.1. This completes the proof of this theorem.