The University of Sydney
MATH2068
Number Theory and Cryptography
(http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/)
Semester 2, 2008
Lecturer: R. Howlett
Tutorial 8
Let
n
be a positive integer. A complex number
ζ
is called an
n
th root of
1 if
ζ
n
= 1.
If in addition
ζ
k
6
= 1 for all positive integers
k
less than
n
then
ζ
is called a
primitive
n
th root of
1.
1.
By factorizing
x
n

1, ﬁnd all the
n
th roots of 1 for
n
= 1,
n
= 2,
n
= 3,
n
= 4
and
n
= 6. In each case determine which of the
n
th roots of 1 are primitive
n
th roots of 1.
Solution.
Obviously the only solution of
x
1

1 = 0 is
x
= 1; so 1 is the only 1st root of 1.
And there are no positive integers less than 1 satisfying 1
k
= 1, because there
are no positive integers less than 1 at all; hence 1 is a primitive 1st root of 1.
Since
x
2

1 = (
x

1)(
x
+ 1) we see that 1 and

1 are the only 2nd roots of 1.
Now (

1)
k
6
= 1 for all positive integers
k <
2, since (

1)
1
6
= 1, and so

1 is a
primitive 2nd root of 1. But 1
1
= 1; so 1 is not a primitive 2nd root of 1. (In
general,
ζ
can only be a primitive
n
th root of 1 for one value of
n
, and when
ζ
= 1 that value is 1, not 2.)
Since
x
3

1 = (
x

1)(
x
2
+
x
+1) we see that the 3rd roots of 1 are 1 and the two
complex solutions of
x
2
+
x
+1 = 0, namely
ω
1
=

1
2
+
√
3
2
i
and
ω
2
=

1
2

√
3
2
i
.
We know that 1 is not a primitive 3rd root of 1 since it is a 1st root of 1;
moreover,
ω
1
and
ω
2
are not 1st or 2nd roots of 1 since (as we have seen) the
only 1st and 2nd roots of 1 are
±
1. So
ω
1
and
ω
2
are primitive 3rd roots of 1.
Since
x
4

1 = (
x

1)(
x
+ 1)(
x
2
+ 1) we see that the 4th roots of 1 are
±
1
and
±
i
. Since
i
and

i
are not 1st, 2nd or 3rd roots of 1 they are primitive 4th
roots of 1. The other two are 2nd roots of 1, hence not primitive 4th roots of 1.
Since
x
6

1 = (
x
3

1)(
x
3
+ 1) = (
x

1)(
x
2
+
x
+ 1)(
x
+ 1)(
x
2

x
+ 1), the
6th roots of 1 are the roots of the polynomials
x

1,
x
+ 1,
x
2
+
x
+ 1 and
x
2

x
+ 1. The ﬁrst three of these give, respectively, the primitive 1st, 2nd and
3rd roots of 1, namely 1,

1,
ω
1
, and
ω
2
. These are the 6th roots of 1 that
are not primitive 6th roots of 1. The two complex roots of
x
2

x
= 1 = 0 are
ζ
1
=
1
2
+
√
3
2
i
and
ζ
2
=
1
2

√
3
2
i
, and we know that these are not 1st, 2nd, 3rd
or 4th roots of 1 (since we have found all of those and they did not include