关于丢番图方程x(x+1)(x+2)=2py^n

摘要

设p为奇素数.证明了:①若整数n>2,则丢番图方程x(x+1)(x+2)=2pyn仅有正整数解(p,x,y)=(3,1,1);②若整数n=2,则丢番图方程x(x+1)(x+2)=2pyn在p■1(mod 8)时仅有正整数解(p,x,y)=(3,1,1),(3,2,2),(3,48,140),(11,98,210);在p≡1(mod 8)时的正整数解为(p,xn,yn)=(p,16t2n,4untnsn),这里p,un,tn,sn满足sn+2=6sn+1-sn,s1=3,s2=17,tn+2=6tn+1-tn,t1=1,t2=6及pu2n=16t2n+1.