Solutions to Assignment 1
September 9, 2009
Section 1.2
1a. Apply Euclidean Algorithm:
7469
=
3
×
2464 + 77
2464
=
32
×
77
So (7469
,
2464) = 77.
3a Note that 423
/
9 = 47
,
198
/
9 = 22. So it suffice to find integer solution to
the equation
47
x
+ 22
y
= 1
Apply Euclidean Algorithm:
47
=
2
×
22 + 3
22
=
7
×
3 + 1
3
=
3
×
1
Working back (or you can also use matrices):
1
=
22

7
×
3
=
22

7
×
(47

2
×
22)
=

7
×
47 + 15
×
22
So a solution is
x
=

7
, y
= 15 (you may obtain other solutions).
14. Any odd number
n
is of the form 4
k
+ 1, or 4
k
+ 3. If
n
= 4
k
+ 1, then
n
2
= (4
k
+ 1)
2
= 16
k
2
+ 8
k
+ 1 = 8(2
k
2
+
k
) + 1
Similarly, if
n
= 4
k
+ 3, then
n
2
= 16
k
2
+ 24
k
+ 9 = 8(2
k
2
+ 3
k
+ 1) + 1
So 8

(
n
2

1) in all cases.
1
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15. Write
x
= 2
m
+ 1
, y
= 2
n
+ 1. Then
x
2
+
y
2
= (2
m
+ 1)
2
+ (2
n
+ 1)
2
=
(4
m
2
+ 4
m
+ 1) + (4
n
2
+ 4
n
+ 1)
=
4(
m
2
+
n
2
+
m
+
n
) + 2
So
x
2 +
y
2
is even, but not divisible by four. In fact, it’s an integer of the form
4
k
+ 2.
Section 1.3
24. Argue by contradiction. Suppose
n
is composite, and all the prime factors
of
n
are
> n
1
/
2
. Write
n
=
ab
, with
a, b
6
= 1. The prime factors of
a, b
are those
of
n
, hence are
> n
1
/
2
. In particular,
a, b > n
1
/
2
. But then
n
=
ab > n
1
/
2
×
n
1
/
2
=
n
a contradiction. Hence some prime factor of
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 Spring '07
 MOK
 Math, Prime number, Equals sign, Divisor, Integer factorization

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