节点文献
关于不定方程组y~2-(k~2+1)x~2=k~2,z~2-((k+1)~2+1)x~2=(k+1)~2
On the Coupled Indefinite Equations y~2-(k~2+1)x~2=k~2,z~2-((k+1)~2+1)x~2=(k+1)~2
【摘要】 证明了标题所述的不定方程组k=7或k=8时仅有x=0的整数解,从而证明了不存在异于1的正整数N,使1,50,65,N和1,65,82,N的任两数之乘积减去1后均为平方数,所用的方法属于初等的。
【Abstract】 The coupled indefinite equations y2 - (k2 + 1 )x2 = k2 and z2 - ((k + 1 )2 + 1 ) x2 = (k + 1 ) 2are proved. When k equal to 7 or 8, their integral solution is only x= 0. Which has proved that the positive integral N except 1 does not exist. In such a condition any poduct between 1,5o, 65, N and 1, 65, 82, N minus 1 is a square numbers. Such a solution can be reached by an elementary method.
【关键词】 丢番图方程;
整数解;
P-1集;
扩张;
二次剩余;
【Key words】 Diophantine equations; Integral solution; P-1 sets; Extension; Quadratic residue;
【Key words】 Diophantine equations; Integral solution; P-1 sets; Extension; Quadratic residue;
- 【文献出处】 浙江工业大学学报 ,Journal of Zhejiang University of Technology , 编辑部邮箱 ,1997年02期
- 【分类号】O156.1
- 【下载频次】15