【摘要】 设a,b,c,x,y是给定的正整数,min(x,y)>1,gcd(ax,by)=1。运用初等数论方法证明了:方程axm+byn=c至多有2组正整数解(m,n);而且当该方程恰有2组正整数解(m,n)=(m1,n1)和(m2,n2)时,必有(axm1-byn1)(axm1-byn1)<0。更多还原

【Abstract】 Let a,b,c,x,y be fixed positive integers such that min(x,y)>1 and gcd(ax,by,)=1.ln this paper,using some elementary number theory methods,we prove that the equation axm+byn=c has at most two positive integer solutions(m,n).Moreover,if the equation has exactly two solutions(m,n)=(m1,n1) and(m2,n2),then we have(axm1-byn1)(axm1-byn1)<0.更多还原