18.781, Fall 2007 Problem Set 8
Solutions to Selected Problems
Problem 2
First, observe the following statement.
If a prime integer (in
Z
)
p
is of the form 3
k
+ 2,
p
is a prime element in
R
.
This can be proved easily using the problem 6 in problem set 7. (Note that if
a
is a prime
factor of
p
,then
N
(
a
) should be
p
. ) By this, we can conclude that 2
,
5
,
11 are still primes in
R
.
If
x
=
r
+
sω
(
r, s
∈
Z
),by easy computation, we have
x
¯
x
=
N
(
x
). It can help us to find the
factorization of the prime integer (in
Z
)
p
. Once we find
r, s
such that
r
2

rs
+
s
2
=
p
, then
we have (
r
+
sω
)(
r
+
s
¯
ω
) = (
r
+
sω
)(
r

s

sω
) =
p
. (Each factor has prime integer value by
the function
N
, so they are primes in
R
.)
By above observation, 2
2

2
·
1+1
2
= 3, 3
2

3
·
1+1
2
= 7 and 3
2

3
·
4+4
2
= 13 implies that
3 = (2+
ω
)(1

ω
), 7 = (3+
ω
)(2

ω
) and 13 = (3+4
ω
)(

1

4
ω
) can be the prime factorization.
In conclusion, we can have following prime factorizations.
7 = (3 +
ω
)(2

ω
)
21 = 3
·
7 = (2 +
ω
)(1

ω
)(3 +
ω
)(2

ω
)
45 = 3
2
·
5 = (2 +
ω
)
2
(1

ω
)
2
·
5
22 = 2
·
11
143 = 11
·
13 = 11
·
(3 + 4
ω
)(

1

4
ω
)
Problem 3
First, prove the following claim.
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 Spring '09
 BRUBAKER
 Number Theory, Prime number, prime integer, R/3R, cubic residue mod, R/5R

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