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Question - Math 324 Assignment 4 (due October 8)

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1. ( 5)(1.5, #46) Show that the nth Fibonacci number fn is divisible by 4 if, and only if,
6 divides n. Hint: Consider the sequence of least residues of fn mod 4.

2. (5)(3.5, #34) Find all pairs of positive integers a, b so that (a, b) = 18 and [a, b] = 504.

3. (6) Determine all solutions x mod 147 of 119x ≡ 98 mod 147.

4. ( 5) Let p be an odd prime and k ≥ 1. What are the solutions mod pk of the congruence
x2 ≡ 1 mod pk ? Prove this.

5. ( 3 + 3)(Continuing Problem 3.7)
a) Show that every non-unit of A can be written as a product of primes of A.
b) ...Read More Is prime factorization in A unique, up to order and associates? Explain why.

6. (5) Let τ(n) be the number of positive divisors of the integer n ≥ 1. Show that τ(n) is a
multiplicative function of n, by using Lemma 3.7.

7. ( 4 + 4) Suppose that m is odd. Let S: = {r1, r2, …, rφ(m)} be a reduced residue system
mod m.
Show that r1r2 … rφ (m) ≡ (− 1) N/2 mod m,
where N is the number of solutions mod m of x2 ≡ 1 mod m. For this, note that S is the
disjoint union of S
0:= {s ∈ S: s2 not ≡ 1 mod m} and S1:= {s ∈ S: s2 ≡ 1 mod m}, and
show that
a) the product of all elements of S0 is ≡ 1 mod m (Hint: Compare proof of Theorem 6.1),
b) the product of all elements of S1 is ≡ ( − 1) N/2 mod m (Hint: For every s ∈ S1, − s is
also in S
1, and s( − s) ≡ − 1 mod m). ...Read Less

Solution Preview - starts as 0, 1, 1, 2, 3, 1, 0, 1 for 0 ≤ n ≤ 7, which is long enough to observe that f n + 6 ≡ f n mod 4 for n = 0, 1, hence for all n ≥ 0 by induction on n: for n ≥ 2, we have f n + 6 = f n − 1 + 6 + f n − 2 + 6 ≡ f n − 1 + f n − 2 = f n mod 4. For 0 ≤ r < 6, another induction, started by the one above, gives f 6q + r ≡

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