证明：有无穷个正整数n，满足n|（2^n）+2. 如果是归纳法，也请详细一点，谢谢！

Question

如题所述

Answer 1

考虑递推数列a[n]: 满足a[n+1] = 2^(a[n])+2, a[1] = 2.
易见数列各项均为偶数, 且严格递增.
用数学归纳法证明a[n] | 2^(a[n])+2对任意正整数n成立.
为证明需要, 证明加强的命题: a[n] | 2^(a[n])+2且a[n]-1 | 2^(a[n])+1, 对任意正整数n成立.

n = 1时, 显然成立.
假设n = k时结论成立, 即a[k] | 2^(a[k])+2 = a[k+1], 且a[k]-1 | 2^(a[k])+1 = a[k+1]-1.
设a[k+1] = b·a[k], a[k+1]-1 = c·(a[k]-1).

由a[k] > 1, a[k+1] = 2^(a[k])+2不是4的倍数, 又a[k]为偶数, 故b为奇数.
因此由2^(a[k]) ≡ -1 (mod 2^(a[k])+1), 有2^(a[k+1]) = (2^(a[k]))^b ≡ (-1)^b = -1 (mod 2^(a[k])+1).
即a[k+1]-1 = 2^(a[k])+1 | 2^(a[k+1])+1.

又由a[k+1]-1为奇数, 可得c为奇数.
因此由2^(a[k]-1) ≡ -1 (mod 2^(a[k]-1)+1),
有2^(a[k+1]-1) = (2^(a[k]-1))^c ≡ (-1)^c = -1 (mod 2^(a[k]-1)+1).
即2^(a[k]-1)+1 | 2^(a[k+1]-1)+1, 于是a[k+1] = 2^(a[k])+2 | 2^(a[k+1])+2.

综上, n = k+1时结论成立.
由数学归纳法原理, 命题对任意正整数n成立.