关于不定方程组x^2-14y^2=1与y^2-Dz^2=16的公解

On the solutions of the simultaneous Diophantine equations x^2-14y^2=1 and y^2-Dz^2=16

设p 1,p 2,…,p s(1≤s≤4)是互异的奇素数,利用递归数列、Pell方程解的性质证明了当D=2p 1p 2…p s(1≤s≤4)时,不定方程组x^2-14y^2=1与y^2-Dz^2=16的整数解如下:当D=2×449时,方程组仅有解(x,y,z)=(±13455,±3596,±120)以及解(x,y,z)=(±15,±4,0);当D≠2×449时,方程组仅有解(x,y,z)=(±15,±4,0).

Let p 1,p 2,…,p s(1≤s≤4)be distinct odd primes.By using recurrent sequence and properties of the solution of the Pell equation,we concluded that,if D=2p 1p 2…p s(1≤s≤4),then the Diophantine equations x^2-14y^2=1 and y^2-Dz^2=16 lead to following integer solutions:when D=2×449,the equations are only associated with solutions(x,y,z)=(±13455,±3596,±120)and(x,y,z)=(±15,±4,0),when D≠2×449,the equations are only associated with solutions(x,y,z)=(±15,±4,0).