Week 4 Homework Solutions
Chapter 3
Section 1
12.The quadratic mean of a and b is always greater than or equal to the arithmatical mean
of a and b.
To prove we start by observing (a-b)2 S 0 and then by algebra that
a
2
- 2ab + b
2
0
a
2
+ b
2
2ab
2a
2
+ 2b
2
a
2
+ 2ab + b
2
(a
2
+ b
2
)/2
[(a + b)
2
]/4
sqrt ((a
2
+ b
2
)/2)
(a + b)/2
as desired.
20.We wish to show x
2
mod 5 = 1 or 4 when x mod 5
0.
We consider four cases:
Case 1: x mod 5 = 1.
Then x
2
mod 5 = 1.
Case 2: x mod 5 = 2.
Then x
2
mod 5 = 4.
Case 3: x mod 5 = 3.
Then x
2
= 9 so x
2
mod 5 = 4
Case 4: x mod 5 = 4.
Then x
2
= 16 so x
2
mod 5 = 1
In all cases, we have shown x
2
mod 5 = 1 or 4.
30.The product of primes of the form 4k+3 is a number of the form 4k+1:
(4m + 3)(4n + 3) = 16mn + 12m + 12n + 9 = 4k + 1 where k = 4mn + 3m + 3n + 2
38.Let n be such that the sum of its divisors is n + 1.
We note 1 and n are divisors of n
and these sum to n + 1.
Hence there are no other divisors of n so n is prime.
44.Let s = (a
1
)(a
2
) . . . (a
m
) and t = (b
1
)(b
2
) . . . (b
n
) be prime factorizations of s and t. Since
gcf (s,t) = 1, a
i
b
j
for all i, j.
The sum of the factors of s is the sum of all possible
combinations of the a
i
and similarly for t and the b
i
.
The product of these sums would
then be the sum of all possible combinations of the a
i
and b
j
considered as a single
group.
Since the lists don't intersect, this is the same as the sum of the factors of st.
Section 2:
6. A. 10, 7, 4, 1, -2, -5, -8, -11, -14, -17
B. 1, 3, 6, 10, 15, 21, 28, 36, 45, 55
C. 1, 5, 19, 65, 211, 665, 2059, 6305, 19171, 58025
D. 1, 1, 1, 2, 2, 2, 2, 2, 3, 3
E. 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
F. 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023
G. 1, 2, 2, 4, 8, 11, 33, 37, 148, 153
H. 1, 2, 2, 2, 2, 3, 3, 3, 3, 3