【摘要】 设k是大于1的奇数,应用初等数论方法证明了:如果2k-1有适合d≡±3(mod 8)的约数d或者ν2(k-1)是奇数,其中ν2(k-1)表示2在k-1的标准分解式中的次数,那么方程x~2+(2k-1)~m=k~n的正整数解(x,m,n)都满足2|n.由此可知:当k<30时,该方程仅有正整数解(x,m,n)=(k-1,1,2).更多还原

【Abstract】 Let k be an odd integer with k>1.Using some elementary number theory methods,we prove that,if either 2k-1is associated with a divisor d with d≡±3(mod 8)orν2(k-1)is odd,whereν2(k-1)is the order of 2in standard factorization of k-1,then every positive integer solution(x,m,n)of the equation x~2+(2k-1)~m=k~n satisfies 2|n.Thus it can be shown that,if k<30,then the equation is associated with only the positive integer solution(x,m,n)=(k-1,1,2).更多还原