高中數學~~整數問題

a, b互相對稱,可設a>=b(同理可得b>=a情形)
(3b-1)/a為(正)整數,又 (3b-1)/a<= (3b-1)/b <3, 故3b-1=a or 3b-1=2a
case1: a=3b-1
(3a-1)/b=(9b-4)/b= 9- 4/b為整數, so, b=1,2,4, thus (a,b)=(2,1),(5,2),(11,4)
case2: 2a=3b-1
(3a-1)/b為整數, then 2(3a-1)/b=(9b-5)/b=9-5/b為整數, thus b=1,5
then (a,b)=(1,1), (7,5)
故(a, b)=(1,1),(2,1),(1,2),(2,2),(5,2),(5,7),(7,5), (4,11),(11,4)
又a, b均大於2, 故(a,b)=(4,11),(11,4),(5,7),(7,5)
Ans: a之所有可能值和=4+5+7+11=27