Question

若正整数a.b.c.x.y,z满足ax=b+c,by=a+c,cz=a+b,则乘积xyz可能有多少个

Answer 1

axbycz = (b+c)(a+c)(a+b) >= 8abc
所以xyz >= 8
设x>=y>=z
若x>=2 ,y>=2 ,z>=2
那么
2(a+b+c) = ax+by+cz >= 2(a+b+c)
所以此时给出x = y = z = 2

若z = 1
那么
c = a+b
ax = b+c = a+2b =>x = 1 + 2b/a

by = a+c = 2a+b =>y = 1 + 2a/b
x和y都是整数
所以a=b
此时给出x = y = 3

综合得x y z可取值为(2,2,2)和(3,3,1)
乘积可能为8或9