- Preliminaries:
  Review notions of :
   . Sets, power set of a set, Cartesian product of two sets, 
     Relations, Functions (all in chapter 0).

- Prove by induction that for all natural numbers n >= 0:  
       n
      sum i^2 = n(n + 1)(2n + 1)/6
      i=1 
                    n
   . Let S(n) be " sum i^2 = n(n + 1)(2n + 1)/6".
                   i=1

   . We prove that S(n) holds for all integers n >= 1.
   , Base: S(1) is trivially true, since: 1^2 = 1 = 1(1 + 1)(2 + 1)/6
   . Ind. Hyp.: Let k >= 1 be an arbitrary integer and assume that
     S(k) holds. 
   . Ind. Step: We prove S(k + 1) is true:
       k+1                    k
       sum i^2 = (k + 1)^2 + sum i^2
       i=1                   i=1

               = (k + 1)^2 + k(k + 1)(2k + 1)/6      By Ind. Hyp.
               = k^2 + 2k + 1 + (2k^3 + 3k^2 + k)/6
               = (2k^3 + 9k^2 + 13k + 6)/6
               = (k^2 + 3k + 2)(2k + 3)/6
               = (k + 1)(k + 2)(2{k + 1} + 1)/6.
                                                 
- Prove by induction that for all natural numbers x >= 2 and n >= 0,
  x^n - 1 is divisible by x - 1.

  . Let S(n) to be "for all natural numbers x >= 2, x^n - 1 is divisible by 
    x - 1".
  . We prove that S(n) is true for all natural numbers n.
  . Base: x^0 - 1 = 1 - 1 = 0 and 0 is definitly divisible by x - 1.
  . Ind. Hyp.: Assume that for an arbitrary natural number k >= 0, S(k) is 
    true.
  . Ind. Step: We want to prove S(k + 1) holds.
      x^{k+1} - 1 = x * (x^k - 1) + (x - 1)
    By Ind. Hyps the first term on the right-hand-side is divisible by
    (x - 1), and the second term is (x - 1) which is trivially divisible 
    by itself (since x >= 2 and therefore x - 1 =/= 0). Therefore
    Their sum (on the right-hand-side) is divisible by (x - 1), and so
    is x^{k+1} - 1.