关于广义Thue－Mahler方程的解数

设ａ，ｂ是非零整数，ｐ１，…，ｐｒ是不同的素数，Ｐ＝｛±｜ｍ１，…，ｍｒ是非负整数｝．设Ｋ是ｎ（ｎ≥３）次代数数域，α１，…，αｍ∈ｋ（１＜ｍ＜ｎ），△（α１，…，αｍ）是α１，…，αｍ的判别式，ｆ（ｘ１，…，ｘｍ）＝αＮｋ／Ｑ（α１ｘ１＋…＋αｍｘｍ）∈ｚ［ｘ１，…，ｘｍ］．本文证明了：当ｆ（ｘ１，…，ｘｍ）非退化且Ｐｉ△（α１，…，αｍ）（ｉ＝１，…，ｒ）时，方程ｆ（ｘ１，…，ｘｍ）＝ｂｙ，ｘ１，…，ｘｍ∈ｚ，ｇｃｄ（ｘ１，…，ｘｍ）＝１，ｙ∈Ｐ至多有（４Ｓｄ２）（Ｓｄ）组解（ｘ１，…，ｘｍ，ｙ），其中ｄ＝ｎ！，Ｓ＝ｒ＋ω是ｂ的不同素因数的个数，ｈＡ是Ｋ的类数．

On the Number of Solutions of the Generalized Thue-Mahler Equation

Let a, b be non-zero integers. Let p1,…,pr be distinct primes, and let P= {± |m1,…, mr are nonnegative integers}. Further let K be an algebraic number fieldof degree n with n ≥3, and let hk denote the class number of K. For α1,…, αm∈K with1 < m < n, let △(α1,… ,αm) denote the discriminant of α1,…,αm, and f(x1,… , xm) =μNK/Q(α1x1 +… + αmxm) ∈ Zx1,…, xm]. In this paper we protve that if f(x1,…, xm) is non-degenerate and pi + △(α1,…,αm) for i=1,…, r, then the equation f(x1,…, xm) = by has at most (4sd2)2 (sd)6 (sd)6 integer solutions (x1,…,xm) satisfy gcd(x1,…,xm) = 1 andy∈P, where d= n!, s = r + ω(b) and ω(b) is the number of distinct prime factors of b.