丢番图方程X^2-(a^2+4p^(2n))Y^4=-4p^(2n)

On the Diophantine Equation X^2-(a^2+4p^(2n))Y^4=-4p^(2n)

令α,n≥1为整数,p为素数.本文证明了丢番图方程X^2-(a^2+4p^(2n))Y^4=-4p^(2n)以及X^2-(a^2+p^(2n))Y^4=-p^(2n)在一定条件下最多只有两组互素的正整数解(X,Y).

Let a, n ≥ 1 be integers, p a prime. We prove that, under some conditions the Diophantine equations X^2- （a^2 ＋ 4p^2n） y^4 =-4p^2n and X^2- （a^2 ＋p^2n）y^4 =-p^2n have at most two coprime positive integer solutions （X, Y）.