
Exercises 263
which is irreducible (exercise) and has all roots real. Therefore the s
j
can be con-
structed using trisector, ruler, and compass.
Then, for example,
r
1
+ r
5
= s
1
r
1
r
5
= (ζ + ζ
12
)(ζ
5
+ ζ
8
)
= ζ
6
+ ζ
9
+ ζ
4
+ ζ
7
= s
3
so r
1
,r
5
are roots of a quadratic over Q(s
1
,s
2
,s
3
). The same goes for the other pairs
of r
j
. Therefore we can construct the r
j
by ruler and compass from the s
j
. Finally,
we can construct ζ from the r
j
by solving a quadratic, hence by ruler and compass.
An explicit construction can again be found in Gleason (1988) and Conway and
Guy (1996) page 200.
Earlier, I said that the Pierpont primes p = 2
a
3
b
+ 1 form a much richer set
than the Fermat primes. ...