关于丢番图方程[(10k1+2)n-1] [(10k2+3)n-1]=x2的解

Solutions on the Diophantine Equation[ (10k1+2)n-1] [ (10k2+3)n-1]=x2

关键词:  丢番图方程  指数方程  解  二次剩余

Key words:  Diophantine equation  exponential equation  solutions  quadratic residue

摘要:

利用二次剩余的方法,证明丢番图方程(aⁿ-1)(bⁿ-1)=x²在(a,b)=(10k₁+2,10k₂+3)时,k₂满足:(1)k₂≡0,1(mod4),(2)k₂≡11,14(mod16),(3)k₂≡6,19(mod64),则这类丢番图方程没有正整数解.

Abstract:

By using quadratic residue module method, it is proved that the Diophantine equation (aⁿ-1) (bⁿ-1)=x² has no solutions for some cases of k₂, where (a, b)= (10k₁+2, 10k₂+3), k₂≡0, 1 (mod 4), or k₂≡11, 14 (mod 4), or k₂≡6, 19 (mod 64).