设a,b,c是给定的正整数,运用初等数论方法证明了:当a+b^2t-1=c^2,b≡5(mod 24),c是适合c≡-1(mod b^2t)的奇数,其中l是任意正整数时,方程a^x+b^y=c^z仅有正整数解(z,y,z)=(1,2l—1,2).
Leta, b, c be fixed positive integers, this paper, using some elementary number theory methods, proves that if a^x+b^y=c^z, b- 5(mod 24)and c is an odd integer with c≡-1(mod b^2l), where l is an arbitrary positive integer, then the equation a^x +b^y =c^z has only the positive integer solution (x, y, z) = (1,2l- 1,2).