By Raphael Pass, Wei-lung Tseng

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**Extra info for A Course in Discrete Structures**

**Example text**

Let us consider another example: b0 = 1; bn = 4bn−1 − 3bn−2 b1 = 1; From the recurrence part of the definition, its looks like the sequence (bn )n will eventually out grow the sequence (an )n . Based only on this intuition, let us conjecture that bn = 3n . Possibly correct proof. Define P (n) to be true if bn = 3n . Base case: P (0) is true since b0 = 1 = 30 . Inductive Step: Assume P (k) is true for 0 ≤ k ≤ n; we wish to show that P (n + 1) is true as well for n + 1 ≥ 3. We have bn+1 = 4bn − 3bn−1 = 4 · 3n − 3 · 3n−1 n+1 = (3 n by P (n) and P (n − 1) n n+1 +3 )−3 =3 Wow!

P − 1 are all relatively prime to p. Therefore φ(p) = p − 1. If n = pq with p = q both prime, then among the numbers 1, 2, . . , n = pq, there are exactly q multiples of p (they are p, 2p, . . , n = qp). Similarly, there are exactly p multiples of q. Observe that that other than multiples of p or q, the rest of the numbers are relatively prime to n; also observe that we have counted n = pq twice. 36. Given a, n ∈ N+ , if gcd(a, n) = 1, then aφ(n) ≡ 1 (mod n) Proof. Let X = {x | x ∈ N + , x ≤ n, gcd(x, n) = 1}.

In the case that p | an+1 we are immediately done (j = n + 1). 31, p | ni=1 ai . We can then use the induction hypothesis to show that there exists 1 ≤ j ≤ n such that p|aj . 33 (Fundamental Theorem of Arithmetic). , the unique prime factorization. For example, 300 = 2 × 2 × 3 × 5 × 5. Proof. We proceed by induction. Let P (n) be the statement “n has a unique prime factorization”. Base case: P (2) is trivial, with the unique factorization 2 = 2. 3. PRIMES 51 Inductive Step: Assume P (k) holds for all 2 ≤ k ≤ n − 1.