Let x be any positive integer greater than 1 , which is not divisible by 17.

Prove that either x^8-1, or x^8+1, is divisible by 17.

let the number be x=17k+r with 1<=r<=16
then (17k+r)^8 mod 17 is congruent to r^8 mod 17
now for r=1 thru 16, r^8 mod 17 is either 1 or 16 (easily verified by direct calculation)
r  r^8 mod 17
1 1
2 1
3 16
4 1
5 16
6 16
7 16
8 1
9 1
10 16
11 16
12 16
13 1
14 16
15 1
16 1

thus in every case either r^8-1 or r^8-1 is congruent to 0 mod 17, thus in every case either x^8+1 or x^8-1 is divisible by 17

Posted by Daniel on 2011-09-05 11:35:33