丢番图方程X^2-(a^2+1)Y^4=3-4a

On the Diophantine Equation X^2-(a^2+1)Y^4=3-4a

设a≥2是正整数.本文证明了:当a=2时,方程X^2一(a^2+1)Y^4=3-4a仅有正整数解(X,Y)=(20,3);当a=3时,该方程仅有2组互素的正整数解(X,Y)=(1,1)和(79,5);当a≥4且4a+1非平方数时,该方程最多有4组互素的正整数解(X,Y);当a≥4且4a+1为平方数时,该方程最多有5组互素的正整数解(X,Y).

Let a ≥ 2 be a positive integer.In this paper,we will prove that if a = 2,then the equation X^2-（a^2 ＋ 1）Y^4 = 3- 4a has only one positive integer solution（X,Y） =（20,3）;if a = 3,then the equation has only two coprime positive integer solutions（X,Y） =（1,1）,（79,5）;if a 4 and 4a ＋ 1 is a nonsquare positive integer,then the equation has at most four coprime positive integer solutions（X,Y）;if a ≥ 4and 4a ＋ 1 is a square,then the equation has at most five coprime positive integer solutions（X,Y）.