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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)=2py“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(mod8)时的正整数解为(p,xn,yn)=(p,16tn^2,4untnsn),这里P,un,tn,sn满足sn+2=6sn+1-s,s1=3,S2=17,tn+2=6tn+1-tn