It is not difficult to recognize Salem numbers of degree 6. Since is
reciprocal, we can write
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
: