关于一类超椭圆丢番图方程

On a kind of superelliptic diophantine equations

获得求解超椭圆丢番图方程da2 x2k+a2k-1x2k-1+… +a1x +a0 =dy2 的快速算法 ,这里a ,d ,k∈N ,d无平方因子且ai∈Z(i=0 ,1,2 ,… ,2k - 1) .给出了超椭圆丢番图方程a2 x4 +x3+2 (a2 - 1)x2+x +(a2 - 2 ) =y2 和a2 x4 -x3+2 (a2 +1)x2 -x +(a2 +2 ) =y2

For the Diophantine equationx^4-Dy^2=1,(1)where D>0 and is not a perfect square,we have proved the following theorems inthis paper.Theorem 1.If D=pq,where p,q are distinct primes,Legendre symbol(p/q)=-1,and the fundamental solution ε= x_0+y_0 D^(1/2) of the Pell's equation x^2-Dy^2=1satisfies r|x_0+1,where r≡3(mod4) is a prime number,then (1) has no positiveinteger solutions x,y.Theorem 2.IfD=2pq,p≡q≡5(mod8),and p,q are distinct primes,then (1)has no solutions in positive integers x,y.Corollary of theoreml.IfD=pq,where p,q are distinct primes,and 1)p≡7 (mod8),q≡1(mod4),(P/q)=-1,or 2)p≡3(mod4),q≡1(mod8),(P/q)=-1,then(1) has nopositive integer solutions x,y.