关于丢番图方程aX~2+D~m=p~z(I)

On diophantine equation aX~2+D~m=p~z(I)

摘要: 设 a、D为正整数 ,a非平方数 ,若丢番图方程 a X2 + D2 y+1 =pz,p| / D,p为奇素数 ,有最小解 ( X,2 y+ 1 ,z) =( b,2 α+ 1 ,d) ,2 | d,则除开当 ab2 D2α+1时 ,( X,Dy -α,z,D2 y+1 -a X2 ,λ) =( Tb( Vl+1 V1-pr02 Vl V1) ,T′( Vl+1 V1+ pr02 Vl V1 ) ,r0 l+ r02 ,U2 l+1 ,-1 ) ;或者当 ab2

Abstract: Let a,D∈N,a0,D0,and let a be a square free,if Diophantine equation aX 2+D 2y+1=p 2,(p|/D,p is a odd prime),has smallest positive integral solution (X,2y+1,z)=(b,2α+1,d),2|d,then (i) if (a,D)=(30,91),then this equation only has integer solutions 30+91=11 2,30·243 2+91=11 6; (ii) if(a,D)≠(30,91),then this equation only has one nonnegative integral solution. except when ab 2D 2α+1,(X,D y-α,z,D 2y+1-aX 2,λ)=(Tb(V l+1V 1-p r 02V lV 1),T′(V l+1V 1+p r 02V lV 1),r 0l+r...