丢番图方程[(10k_1+2)~n-1][(10k_2+5)~n-1]=x^2的解

On the Solution to Diophantine equation[(10k_1+2)~n-1] [(10k_2+5)~n-1]=x^2

本文利用二次剩余的方法,讨论了丢番图方程在(a,b)=(10+2,10+5)时的解,解决了当满足某些条件的这一类丢番图方程的解的情况。

In this paper, by using quadratic residue, we have discussed the solution to the Diophantine equation [（10k1＋2）^n-1][（10k2＋5）^n-1]=x^2, hence working out the solutions when k2 meets some certain terms of such Diophantine equations.