数论设m>n≥0,证明(2^（2^n）+1)|(2^（2^m）-1)

Question

数论 设m>n≥0,证明(2^（2^n）+1)|(2^（2^m）-1)

Answer 1

其实就是因式分解.
2^(2^m)-1 = (2^(2^(m-1))-1)(2^(2^(m-1))+1)
= (2^(2^(m-2))-1)(2^(2^(m-2))+1)(2^(2^(m-1))+1)
...
= (2^1-1)(2^1+1)(2^2+1)(2^4+1)...(2^(2^(m-2))+1)(2^(2^(m-1))+1)
= ∏{0 ≤ i ≤ m-1} (2^(2^i)+1)
而由m > n,有n ≤ m-1,故2^(2^n)+1是乘积中的一项.
可知2^(2^n)+1 | 2^(2^m)-1.