求A=根号((9n-1)/(n+7))为有理数的正整数n的值

求A=根号((9n-1)/(n+7))为有理数的正整数n的值
求A=根号((9n-1)/(n+7))为有理数的正整数n的值

求A=根号((9n-1)/(n+7))为有理数的正整数n的值
由n是正整数,√((9n-1)/(n+7))为有理数当且仅当(n+7)·√((9n-1)/(n+7)) = √((9n-1)(n+7))为有理数.
而(9n-1)(n+7)是整数,其平方根若为有理数则必为整数.
设非负整数m满足m² = (9n-1)(n+7) = 9n²+62n-7.
则9m² = (9n)²+62(9n)-63 = (9n+31)²-31²-63 = (9n+31)²-1024.
即有1024 = (9n+31)²-9m² = (9n+31+3m)(9n+31-3m).
9n+31+3m是1024的约数.
n是正整数,m是非负整数,故9n+31+3m是大于31的整数.
此外,易见9n+3m+31除以3余1.
满足条件的1024的约数有64,256,1024.
若9n+31+3m = 64,有9n+31-3m = 1024/64 = 16,解得n = 1.
若9n+31+3m = 256,有9n+31-3m = 1024/64 = 4,解得n = 11.
若9n+31+3m = 1024,有9n+31-3m = 1024/64 = 1,解不为整数.
可验证n = 1时√((9n-1)/(n+7)) = 1,n = 11时√((9n-1)/(n+7)) = 7/3为有理数.
故满足条件的正整数为1,11.

令f(n)=(9n-1)/(n+7)=[9(n+7)-64]/(n+7)=9-[64/(n+7)]
根据定义域,以及题目条件
(9n-1)/(n+7)≥0
n≥1/9
要使A为整数,即f(n)为整数
n+7>7,
n+7=8, n=1
n+7=16 n=9
n+7=32 n=25
n+7=64 n=57