Totally real Thue inequalities over imaginary quadratic fields: an improvement
aa r X i v : . [ m a t h . N T ] F e b Totally real Thue inequalitiesover imaginary quadratic fields: an improvement
István Gaál ∗ , University of Debrecen, Mathematical InstituteH–4002 Debrecen Pf.400., Hungary, e–mail: [email protected] ,Borka Jadrijević
University of Split, Faculty of Science,Ruđera Boškovića 33, 21000 Split, Croatia, e–mail: [email protected]
László Remete † University of Debrecen, Mathematical InstituteH–4002 Debrecen Pf.400., Hungary, e–mail: [email protected]
February 26, 2021
Mathematics Subject Classification: Primary 11D59; Secondary 11D57Key words and phrases: relative Thue equations, Thue inequalities
Abstract
We significantly improve our results of [1] reducing relative Thue inequalities to abso-lute ones.
Let F ( x, y ) be a binary form of degree n ≥ with rational integer coefficients. Assume that f ( x ) = F ( x, has leading coefficient 1 and distinct real roots α , . . . , α n . Let < ε < andlet K ≥ . Let A = min i = j | α i − α j | , B = min i Y j = i | α j − α i | , C = K (1 − ε ) n − B , G = K /n εA . Let m ≥ be a square-free positive integer, and set M = Q ( i √ m ) . Consider the relativeinequality | F ( x, y ) | ≤ K in x, y ∈ Z M . (1) ∗ Research supported in part by the EFOP-3.6.1-16-2016-00022 project. The project is co-financed by theEuropean Union and the European Social Fund. † Research supported by the ÚNKP-19-3 new national excellence program of the Ministry of human capacities. f F is irreducible, then (1) is called a Thue inequality. We emphasize that our statements arevalid also if F is reducible.If m ≡ , then x, y ∈ Z M can be written as x = x + x i √ m x + x ) + x i √ m , y = y + y i √ m y + y ) + y i √ m , and if m ≡ , , then x = x + x i √ m, y = y + y i √ m, in both cases with x , x , y , y ∈ Z . Set s = 2 if m ≡ and s = 1 if m ≡ , . In the following Theorem we formulate our statements parallely in the two cases. Theorem 1
Let ( x, y ) ∈ Z M be a solution of (1). Then | F ( sx + ( s − x , sy + ( s − y ) | ≤ s n K, | F ( x , y ) | ≤ s n K ( √ m ) n , (2) and | F ( sx + ( s − x , sy + ( s − y ) | · | F ( x , y ) | ≤ s n K n · ( √ m ) n . (3) If | y | > max ( G, (cid:18) s · C √ m (cid:19) n − ) , then x y = x y . (4) If | y | > max n G, ( s · C ) n − o and sy + ( s − y = 0 , then sx + ( s − x = 0 . (5) If | y | > max ( G, (cid:18) s · C √ m (cid:19) n − ) and y = 0 , then x = 0 . (6) Remark 1.
The present inequality (2) is much sharper than the corresponding inequalitiesof Theorem 2.1 of [1]. Moreover we obtain these inequalities without any conditions on thevariables. This makes the applications much easier. If the values of F are non-zero, then (3)yields further new restrictions for the possible solutions of (1). Proof of Theorem 1.
Let ( x, y ) ∈ Z M be an arbitrary solution of (1). Let β j = x − α j y, j = 1 , . . . , n , then inequality(1) can be written as | β · · · β n | ≤ K. (7)We have β j = 1 s (( sx + ( s − x ) − α j ( sy + ( s − y )) + i √ ms ( x − α j y ) . Obviously, | Re( β j ) | ≤ | β j | , | Im( β j ) | ≤ | β j | , ≤ j ≤ n. n Y j =1 | Re( β j ) | ≤ n Y j =1 | β j | ≤ K, and n Y j =1 | Im( β j ) | ≤ n Y j =1 | β j | ≤ K, which imply (2). Moreover, n Y j =1 | Re( β j ) |· n Y j =1 | Im( β j ) | = n Y j =1 ( | Re( β j ) | · | Im( β j ) | ) ≤ n Y j =1 | Re( β j ) | + | Im( β j ) | n Y j =1 | β j | ≤ K n , whence we obtain (3).Assume now | y | ≥ G. (8)Let i be the index with | β i | = min j | β j | . Then | β i | ≤ K n and for j = i | β j | ≥ | β j − β i | − | β i | ≥ | α j − α i | · | y | − K n ≥ (1 − ε ) · | α j − α i | · | y | . (9)From (7) and (9) we have | β i | ≤ K Q j = i | β j | ≤ C | y | n − . (10)Using that α i | y | is real, by (10) we obtain | Im( xy ) | = | Im( α i | y | − xy ) | ≤ (cid:12)(cid:12) α i | y | − xy (cid:12)(cid:12) = | y | · (cid:12)(cid:12)(cid:12)(cid:12) α i − xyyy (cid:12)(cid:12)(cid:12)(cid:12) = | y | · (cid:12)(cid:12)(cid:12)(cid:12) α i − xy (cid:12)(cid:12)(cid:12)(cid:12) ≤ C | y | n − . If | y | > (cid:18) s · C √ m (cid:19) n − , then this implies x y = x y .Inequality (10) indicates that | β i | is small for sufficiently large | y | and so are its real andimaginary parts that can equal zero if we impose some extra assumptions.–If | y | > ( sC ) n − , then | ( sx +( s − x ) − α i ( sy +( s − y ) | < . So, sy +( s − y = 0 implies sx + ( s − x = 0 .–If | y | > (cid:18) sC √ m (cid:19) n − , then | x − α i y | < . So, y = 0 implies x = 0 . (cid:3) Finally, we give useful hints for a practical application of Theorem 1. Using the same no-tation let us consider again the relative inequality (1). We describe our algorithm in case m ≡ , since the other case is completely similar.First, we solve F ( x , y ) = k for all k ∈ Z with | k | ≤ n K/ ( √ m ) n . Since the equation F ( x , y ) = 0 can also have non-trivial solutions if F is reducible, we split our arguments intotwo cases. 3. First suppose F ( x , y ) = 0 . This makes possible to determine x , y . If F is irreducible,then x = y = 0 , if F is reducible, then x , y can be determined easily (if there are any). Wethen determine the solutions ( a, b ) ∈ Z of | F ( a, b ) | = k for all k with | k | ≤ n K . Using allpossible values of x , y for each solution ( a, b ) we determine x = ( a − x ) / , y = ( b − y ) / and check if these are integers. (Note that if F is irreducible, then x = y = 0 implies | F ( x , y ) | ≤ K and the procedure can be simplified.) Having all possible x , x , y , y we testif ( x, y ) ∈ Z M is a solution of (1).B. Assume now F ( x , y ) = k = 0 for some ( x , y ) ∈ Z . Then we solve F ( a, b ) = k in ( a, b ) ∈ Z for all k ∈ Z with | k k | ≤ n K / ( √ m ) n (a part of this calculation was alreadyperformed by solving F ( x , y ) = k ). Having a, b, x , y we calculate x = ( a − x ) / , y =( b − y ) / . For x , y and integer values x , y we test if ( x, y ) ∈ Z M is indeed a solution of (1). Remark 2. If m is sufficiently large, then by (2) we have | F ( x , y ) | < . In case F isirreducible, this implies x = y = 0 , whence (1) reduces to an inequality in x , y over Z . Remark 3.
Solving Thue equations over Z is no problem any more by using well-knowncomputer algebra packages. If F is reducible, this task is even easier. References [1] I. Gaál, B. Jadrijević and L. Remete,
Totally real Thue inequalities over imaginaryquadratic fields , Glas. Mat., III. Ser.53