Q13/091) Prove that for every prime p > 7, p^6 - 1 is divisible by 504.

we have 504 = 9 * 56 = 7 * 8 * 9

we need to show that it is divisible by 7 , 8 and 9 as they are pairwise coprime

now as p is prime and 7 is also prime

so p^6 = 1 mod 7 by Fermat's Little theorem so p^6-1 is divisible by 7

p^6-1= (p^3-1)(p^3+1)

as these 2 are 2 successive even numbers Reason p is odd so p^3 is odd

so one is divisible by 4 and another by 2 so product is divisible by 8

one of p^3-1 and p^3 + 1 is divisible by 9

as p > 7 so p mod 3 = 1 or -1

to prove is if p = 1 mod 3 then p = (3n+1)

so p^3 = 27n^3 + 27n^3 + 9n + 1 or p^3-1 = 9(3n^3+3n^2 + 1) or dibsible by 9

if p = 3n-1 then p^3-1= 9(3n^3-3n^2 + 1) or divisible by 9

so p^6 -1 is divisible by 9

hence is divisible by 7 * 8 * 9 or 504