2.Let r be an oddinteger with r > 1, m be an even integer. Let Ur, Vr beintegers satisfying and c was a power of prime, then the equation ax + by = cz had only the positive integer solution (x,y,z) = (2, 2,r).
3.Let r be an oddinteger with r>1.In this paper the author gives a necessary condition for(X,Y,Z) being a positive integer solution of the equation X~2+Y~2=Z~r with Y being a power of an odd prime.
4.Let D be an oddinteger with D>1,and let p be a prime with D■0(mod),this paper givs a necessary and sufficient condition for the equation p2x-pxDy+D2y=z2 to have the nonnegative integer solutions(x,y,z) with y>1.
5.5(mod8), c is a power of a prime and the equationax+by=cz has a positive integer solution (x, y, z) = (2,2, r), where r is an oddinteger with r>1, then the exceptional solutions (x, y, z) of the equation satisfy x=2 and y=z=1(mod 2).
6.Let r be an oddinteger with r>1,and let u,v be positive integers such that 2|u gcd (u,v) and u>2r v/π,Further let a,b,c be positive integers satisfying a+b-1=(u+v-1)~r and c=u~2+v~2.The value of Jacobi symbol (b/v)/(a/u) is determined.
8.The even or odd quality of an integer. If two integers are both odd or both even, they are said to have the same parity; if one is odd and one even, they have different parity.
9.We prove that if a+b2l-1=c2,b ≡ 5(mod 12) and c is an odd prime with c≡-1(mod b2l),where l is a positive integer,then the equation ax+by=cz has only the positive integer solution(x,y,z)=(1,2l-1,2).
本文证明了:当a+b2 l-1=c2,b≡5(m od 12),c是适合c≡-1(m od b2 l)的奇素数,其中l是正整数时,方程ax+by=cz仅有正整数解(x,y,z)=(1,2 l-1,2).收藏指正
10.In this paper, let p be an odd prime with p>3, we prove that the equation (xp-yp)/(x-y)=z2 has only the positive integer solution (x, y, z, p)=(3,1,11,5), satisfying x>y+1. gcd (x, y)=1. As a result, x is an odd prime power.
