期刊文献+

关于费马大定理(Ⅱ) 被引量:2

ON FERMAT'S LAST THEOREM (II)
下载PDF
导出
摘要 证明了方程x^(2p)+y^(2p)=z^2((x,y)=1,P(>3)是素数)如有解,则必有4P^2|x或4P^2|y.对方程x^(2p)+y^2=z^(2p),x^(2p)+y^(2p)=z^p和x^(2p)+y^p=z^(2p)也得到了类似的结果.此外,我们还有以下的结果:(1)设r(N)表示使得方程x^(2n)+y^(2n)=z^2有解的正整数n(≤N)的个数,则r(N)=o(N)(N→∞).(2)如果正整数x,y,z和n满足x^n+y^n=z^n,x<y<z,n>2,则必有x^2>nz+n-3. In this paper, it has been proved that if the equationx2p + y2p=z2 (x, y) =1 and a prime p>3 (1)has integer solution, then 4p2|x or 4 P2|y. For the equationsx2p + y2=z2p, x2p + y2p=z2p and x2p +yp = z2pthe results which are analogous to (1) are also established. In addition, other two results are also obtained:1. Let r(N) be the number of n<N with the equation x2n+y2n=z2 has integer solution. Then r(N)=o(N). as N→∞.2. If the equation xn + yn= zn has a solution in positive integers x<y<z for some n>2, them x2>nz + n-3.
作者 曹珍富
机构地区 哈尔滨工业大学
出处 《西南师范大学学报(自然科学版)》 CAS CSCD 1991年第1期12-17,共6页 Journal of Southwest China Normal University(Natural Science Edition)
关键词 丢番图方程 费马大定理 同余 Diophantos equation Fermat's last theorem congruence inequality
  • 相关文献

参考文献2

共引文献1

同被引文献19

  • 1汤健儿.不定方程 x^3+y^3=z^2与 x^3+y^3=z^4[J].数学的实践与认识,1993,23(1):90-94. 被引量:32
  • 2孙琦 曹珍富.关于丢番图方程X^p-Y^p=Z^2[J].数学年刊:A辑,1986,7(5):514-518.
  • 3曹珍富.关于费尔马大定理(I)[J].西南师范大学学报,1988,(4):15-19.
  • 4曹珍富.关于丢番图方程x^p-y^p=Dz^2[J].东北数学,1986,2(2):219-227.
  • 5Powell B. sur L′ equation Diophantine x4± y4= zp[J]. Bull SC Math,1983,107:219-223.
  • 6Darmon H. The equation x4 -y4=zp[J]. C R Math Rep. Acad SCI Canada,1993,15(6) :286-290.
  • 7K. Wu,M. Le. A note on the Diophantine equation x4 - y4 =zp[J].C. R Math Rep. Acad SCI Canada, 1995, (17), 197 - 200.
  • 8Darmon H. Granville A. on the equations zm= F(x, y) and Axp+Byq=Czr[J]. Bull London Math Soc,1995, (27) :513-543.
  • 9R. D. Mauldin. A generalization of fermat′s last theorem: the Beal coniecture and prize problem, Notices of the Amer. Math. Soc.1997.11(44) :1436-1437.
  • 10孙琦 曹珍富.关于丢番图方程xp—yp=z2[J].数学年刊:A辑,1986,7(5):514-518.

引证文献2

二级引证文献45

相关作者

内容加载中请稍等...

相关机构

内容加载中请稍等...

相关主题

内容加载中请稍等...

浏览历史

内容加载中请稍等...
;
使用帮助 返回顶部