中文摘要：

a，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）．

英文摘要：

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）.