| 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. |
| 关键词: 丢番图方程 指数方程 解 二次剩余 |
| DOI： |
| 投稿时间：2007-01-19 |
| 基金项目:四川省教育厅自然科学基金项目(2006C057);阿坝师专校级科研基金项目资助 |
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| Solutions on the Diophantine Equation[ (10k1+2)n-1] [ (10k2+3)n-1]=x2 |
|
TANG Bo, YANG Shi-chun
|
| (Department of Mathematics, Aba Teachers College, Wenchuan, Sichuan, 623000, China) |
| 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). |
| Key words: Diophantine equation exponential equation solutions quadratic residue |
