摘要:
设a是正整数.本文证明了:当a=1时,方程X2-(a2+1)Y4=8-6a仅有正整数解(X,Y)=(2,1);当a=2时,该方程仅有正整数解(X,Y)=(1,1);当a=3时,该方程无正整数解(X,Y);当a=4时,该方程仅有2组互素的正整数解(X,Y)=(1,1)和(103,5);当a≥5且6a+1非平方数时,该方程最多有3组互素的正整数解(X,Y);当a≥5且6a+1为平方数时,该方程最多有4组互素的正整数解(X,Y).%Let a be a positive integer.In this paper,we prove that if a =1,then the equation X2-(a2 + 1)Y4 =8-6a has only one positive integer solution (X,Y) =(2,1);if a =2,then the equation has only one positive integer solution (X,Y) =(1,1);if a =3,then the equation has no positive integer solution (X,Y);if a =4,then the equation has only two coprime positive integer solutions (X,Y) =(1,1),(103,5);if a ≥ 5 and 6a + 1 is a nonsquare positive integer,then the equation has at most three coprime positive integer solutions (X,Y);if a ≥ 5 and 6a + 1 is a square,then the equation has at most four coprime positive integer solutions (X,Y).