关于丢番图方程px^4-(p-1)y^2=z^4

On Diophantine equation px^4-(p-1)y^2=z^4

对任意的奇素数p,还没有找到给出丢番图方程px4-(p-1)y2=z4的全部正整数解的统一的初等方法,目前只解决了某类特殊的奇素数p的求解问题,例如王洪昌等人完全解决了p-1=Q2;或2Q2;或qQ2,2|Q,q≡3(mod4)为奇素数,Q为正整数的情形.认为对某类特殊的奇素数p求解丢番图方程px4-(p-1)y2=z4,目的是对任意的奇素数p,寻找给出丢番图方程px4-(p-1)y2=z4的全部正整数解的统一解法.当p=2q+1,q≡5(mod8),p,q为奇素数时,利用初等方法把方程px4-(p-1)y2=z4化为方程x2+my2=z2,从而给出方程px4-(p-1)y2=z4的全部正整数解;当q为任意正整数时,上述解法仍然适用,因此对任意给定的奇素数p,实际上已经给出了丢番图方程px4-(p-1)y2=z4的全部正整数解的统一解法.

When p is an arbitrary odd prime number,there is no unified primary method of all positive integer solutions to the Diophantine equation px4-（p-1）y2=z4,but only the solution of some certain types of special odd primes number p,such as Wang Hongchang etc have completely resolved p-1=Q2;or 2Q2;orqQ2,2｜Q,q≡3（mod4） with q is an odd prime number and Q is a positive integer.People solve the Diophantine equation px4-（p-1）y2=z4 with some special types of odd prime number p to find a unified approach of all positive integer solutions to Diophantine equation px4-（p-1）y2=z4.When p=2q＋1,q≡5（mod8）,p,q is an odd prime number,we put equation px4-（p-1）y2=z4 into equation x2＋my2=z2 by the elementary methods,by which we give the all solution to Diophantine equation px4-（p-1）y2=z4.When q is an arbitrary positive integer,the above solution is still applicable.In fact,we have given a unified method of all positive integer solutions to Diophantine equations px4-（p-1）y2=z4,when p is an arbitrary given odd prime number.