of degree 6 has a minimal polynomial of the following form:
where a,b and c are integers. The trace of
is
, where the sum is over the six conjugates
of
, i.e., the six roots of (2.1). Two of these conjugates are
and
and the remaining conjugates satisfy
, so a bound on |a| implies
a bound on
and hence on |b| and |c|. So, for fixed a, there is a finite
set of
with
. Observe that
so
for all Salem numbers of degree 6. In fact, there are no
such numbers with trace -1, 4 with trace 0, 15 with trace 1, and 39 with
trace 2.
It is not difficult to recognize Salem numbers of degree 6. Since
is
reciprocal, we can write
where
The zeros of U are the numbers
and hence U must have two
roots in the open interval -2 < x < 2 and one root with x > 2. This is equivalent
to the following requirements: (i)
, (ii)
has real roots,
the smaller of which,
satisfies
, and
. We used
this criterion in compiling our list of Salem numbers.
We also require U to be irreducible. Since U is cubic, it suffices that
for any integer n. Only a finite set of n need be checked since, by a well-known
estimate of Cauchy,
for
. It is easy
to see that P can only factor into factors of even degree since the roots
and
must belong to the same factor, or else there would be a factor the product of
whose roots is in absolute value less than 1, which is clearly impossible. Thus, P
is irreducible if and only if U is irreducible.
The following useful result is an elementary deduction from the above discussion.
Let P be as in (2.1) and U as in (2.3). A necessary condition for P to be the minimal polynomial of a Salem number is that![]()
for
. A sufficient condition for P to be the minimal polynomial of a Salem number is, in addition to (2.4), any one of
,
or
.
In general, if P is the minimal polynomial of a Salem number of degree d = 2s, then
there is a monic polynomial U with integer coefficients for which
. The numbers
are the zeros of U repeated twice each. The
following computation gives a useful expression for
:
