1st of MI Question

Let P(n) be the statement:
(x + 1) is a factor of xn + 1 where n is an odd integer.
When n = 1,
xn + 1 = x + 1 which is divisible by (x + 1)
So P(1) is true.
Now, suppose P(2k + 1) is true, where r is a positive integer:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep08/CrazyMI1.jpg

Hence, P(2k + 3) is also true.
By the principle of MI, P(n) is true for all positive integer n.

If n is a negative odd integer, then we let m = -n, where m > 0
So, from above, (x + 1) is a factor of x^m + 1
(x + 1) is a factor of x^(-n) + 1

2008-09-04 21:13:30 補充：
Here I denote x^(-n) + 1 = M(x)(x + 1), where M(x) is a polynomial in terms of x.
So, x^(-n) + 1 = x^(-n)[1 + x^n] = x^(-n)M(x)(x + 1) = Q(x)(x + 1), where Q(x) = x^(-n)M(x)
That is, (x + 1) is a factor of x^n + 1.
This is also valid when n is a negative odd number.