BRIEF ANSWERS FOR TEST 1

• T or F?
  • T. Any subset of a well-ordered set, under the induced ordering, is well-ordered. (Reason: Assume P is a subset of a well ordered set N. Any nonempty subset S of P is a nonempty subset of N, and therefore has a least element. Thus any nonempty subset of P has a least element.)
  • F. Well-ordered sets are linearly ordered, and the collection of ideals of Z under inclusion is not. In particular, the set {2*Z,3*Z} has no least element.
  • F. For example, if m = a = n = b = 2, then d = 8. But 8 is not equal to gcd(a,b) = 2.
  • F. The prime divisors of (1000-choose-500) are all divisors of 1000!, hence they are less than 1000.
  • F. gcd(28,63) = 7, which does not divide 12. Hence there is no solution.
• This is a problem where you must give a proof by induction of the statement:

  Sn: F_n = ((n-1)-choose-0) + ((n-2)-choose-1) + ...

  Your argument should involve the following features:

  • TWO consecutive statements in the sequence must be verified in the basis step in order to use the relation F_(n+1) = F_n + F_(n-1) in the inductive step. (So verify S1 and S2.)
  • In the inductive step, the cases where n is even and where n is odd require SEPARATE arguments.
  • The key to success in the inductive step is the identity

    ((n+1)-choose-(k+1)) = (n-choose-k) + (n-choose-(k+1)).

• You can solve this using the Euclidean algorithm, by factoring, or by asking Maple to solve the equation 474*x-1213*y=1. (Use the isolve command.) With the latter method you find that 474(888)-1213(374)=1. This proves that x = 888 (mod 1213) is the solution to 474*x = 1 (mod 1213). The fact that a solution exists implies that gcd(474,1213) = 1.
• Assume that m = n^2 and that m = p_1^(e_1) * * * p_k^(e_k). If the prime factorization of n is q_1^(f_1) * * * q_l^(f_l), then

  p_1^(e_1) * * * p_k^(e_k) = m = n^2 = q_1^(2f_1) * * * q_l^(2f_l).

  By uniqueness of prime factorizations, the q_i's must be a reordering of the p_j's and the exponents must agree. We may assume that p_i = q_i for all i, and so f_i = e_i/2. Hence the square root n of m has the factorization

  n = q_1^(f_1) * * * q_l^(f_l) = p_1^(e_1/2) * * * p_k^(e_k/2).

• By Fermat's Theorem we have q^p = q (mod p). We can cancel q from both sides since gcd(q,p) = 1, so q^(p-1) = 1 (mod p). Since we obviously have p^(q-1) = 0 (mod p) we get

  p^(q-1) + q^(p-1) = 1 (mod p).

  Similarly we have

  p^(q-1) + q^(p-1) = 1 (mod q).

  Therefore, since (p^(q-1) + q^(p-1) - 1) is divisible by both p and q it is divisible by lcm(p,q) = pq. This yields

  p^(q-1) + q^(p-1) = 1 (mod pq).

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  Last modified on Oct 1, 1998.