关于丢番图方程x^4+Dy^4=z^2

On the Diophantine Equation x^4+Dy^4=z^2

设p为奇素数.利用同余性质及Fermat的无穷递降法,证明了:D=p3,p≡3,7(mod 16);或D=-p3,p≡9,13(mod 16);或D=2p3,p≡3,5(mod 8);或D=4p3,p≡3,7(mod 16)时,方程x4+Dy4=z2,gcd(x,y)=1均无正整数解.同时给出D=3时方程的全部正整数解.

Let p be an odd prime.Using the properties of congruence and Fermat method of infinite descent,it is proved that if D=p3,p≡3,7（mod 16） or D=-p3,p≡9,13（mod 16） or D=2p3,p≡3,5 （mod 8） or D=4p3,p≡3,7（mod 16）,then the Diophantine equation x4 ＋Dy4 =z2 has no positive integer solutions x,y,z with gcd（x,y）=1.Additionally,all positive integer solutions of the equation when D=3 are given.