Q13/122) If k is a positive integer and x is an integer such that x - 2 is a multiple of 7 and x^6 - 1 is a multiple of 7^k, show that (x + 1)^6 - 1 is also a multiple of 7^k.

We have 

(a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + y^6

= a(a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) + y^ 6

If x -2 is divisible by 7 then taking a = 7k and b = 2

=> x = 7k+ 2 , b^6 mod 7 = 1

(a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) mod 7 = 6b^5 mod 7 not divisible by 7

now if a is divisible by 7^m and so (a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) is not divsible by 7

so x^6 – 1 mod 7^m is zero but x^6 – 1 mod 7^(m+1) is not zero

(x+1)^6 – 3^ 6 (as x+1 = 7k + 3) mod 7^(m+1) is is not zero

so if x^6 – 1 is divisible by 7^m <=> (x+1)^6 -1 is is divisible by 7^m

  I have proved that 
x^6 - 1 is a multiple of 7^k, iff (x + 1)^6 - 1 is also a multiple of 7^k.