摘要
设p是模8余1的素数.该文用柯召-Terjanian-Rotkiewicz方法和高次丢番图方程的有关结果证明了,如果丢番图方程2py^(2)=x(x+1)(x+2)有正整数解,则p满足py^(2/1)=s2rn,其中n是奇数,且当n>1时其每个奇素因子q都使得(-pq)=1.这里s 2 rn+t2rn√2=(3+2√2)^(2 rn).还给出了关于丢番图方程p^(2)x^(4)-dy^(2)=1的一个有趣的结果.
Let p be an odd prime congruent to 1 modulo 8.In this paper we prove,by Ko Zhao-Terjanian-Rotkiewicz method and relevant results about Diophantine equations of higher order,that if the Diophantine equation 2py^(2)=x(x+1)(x+2)has a positive integer solution,then p satisfies py^(2/1)=s2rn,where n is an odd positive integer,and each odd prime factor of n satisfies-p q=1 when n>1.Here s 2 rn+t2rn√2=(3+2√2)^(2rn).We also give an interesting result about the Diophantine equation p^(2)x^(4)-dy^(2)=1.
作者
何宗友
HE Zong-you(Shenzhen Jingtian Precision Technology Co.,Ltd.,Shenzhen 518118,China)
出处
《南宁师范大学学报（自然科学版）》
2023年第4期6-9,共4页
Journal of Nanning Normal University：Natural Science Edition
