关于不定方程3x(x+1)(x+2)(x+3)=26y(y+1)(y+2)(y+3)

On the Diophantine equation 3x(x+1)(x+2)(x+3)=26y(y+1)(y+2)(y+3)

运用递归序列和平方剩余的方法,证明了不定方程3x(x+1)(x+2)(x+3)=26y(y+1)(y+2)(y+3)仅有正整数解(x,y)=(13,7)。

In this paper,with the primary methods of recurrence sequences and quadratic remainders,the author shows that the Diophantine equation 3x(x+1)(x+2)(x+3)=26y(y+1)(y+2)(y+3)has a unique positive integer(x,y)=(13,7).