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18.781, Fall 2007 Problem Set 10
Solutions to Selected Problems
Problem 1
We have the following identity :
sin
2
(2
θ
) = 4
sin
2
θcos
2
θ
= 4
sin
2
θ
(1

sin
2
θ
)
Applying
θ
=
π
12
and let
α
=
sin
(
π
12
)
.
Using
sin
π
6
=
1
2
, we have
1
4
= 4
α
2
(1

α
2
)
.
That is,
16
α
4

16
α
2
+ 1 = 0
.
Therefore,
α
=
sin
(
π
12
) is algebraic.
2
Problem 2
(a) First, let’s prove following claim.
There is no element
α
∈
Z
[
√

5] such that
N
(
α
) = 2 or 3.
To prove this, it is enough to show that there is no integral solution (
x, y
) satisfying
x
2
+5
y
2
=
2 or 3. This is clear.
For
α
= 2
,
3
,
1 +
√

5
,
1

√

5, we have
N
(2) = 4
, N
(3) = 9
, N
(1 +
√

5) =
N
(1

√

5) = 6
.
Thus for each
α
, if there is any further factorization
α
=
βγ
with
N
(
β
)
, N
(
γ
)
>
1(i.e. each of
β, γ
is not a unit), since
N
(
α
) =
N
(
β
)
N
(
γ
),
N
(
β
) or
N
(
γ
) should be 2 or 3, which is absurd
by the claim.
Hence, 6 is not factorized uniquely, and this implies that
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 Spring '09
 BRUBAKER
 Number Theory

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