Suppose
$$ P(x)=x^mQ(x+x^{-1}) $$
is primitive, irreducible, reciprocal, monic, noncyclotomic, and integral. Assume \(Q\) is totally real and has exactly two roots outside \([-2,2]\).
If the product of the two roots of \(P\) outside the unit circle is nonreciprocal, then
$$ \boxed{M(P)>2.} \tag{1} $$
So every polynomial in this class below two has a reciprocal Mahler product. In particular, any candidate below Lehmer's number is forced into that arithmetic branch.
This is not a general lower bound for reciprocal polynomials. The assumptions "totally real trace" and "exactly two exterior pairs" are doing real work. What delights me is how little is needed after them: one endpoint integer and one Galois-orbit degree collapse.
The endpoint integer#
Let the two exterior trace roots come from reciprocal-root moduli \(A,B>1\), and put
$$ M=AB=M(P). $$
For either sign of a trace root,
$$ |u^2-4|=(A-A^{-1})^2. $$
The product of the two exterior factors satisfies
$$ \begin{aligned} (A-A^{-1})(B-B^{-1}) &=M+M^{-1}-\left(\frac AB+\frac BA\right)\\ &\le M+M^{-1}-2, \end{aligned} \tag{2} $$
with equality exactly when \(A=B\).
Every interior trace root \(w\in[-2,2]\) contributes
$$ |4-w^2|\le4. $$
Because \(P\) is irreducible and noncyclotomic, neither endpoint vanishes:
$$ Q(2)Q(-2)\in\mathbb Z\setminus\{0\}. $$
Multiplying all trace-root factors gives
$$ 1 \le |Q(2)Q(-2)| \le 4^{m-2}\bigl(M+M^{-1}-2\bigr)^2. \tag{3} $$
The left side is the whole arithmetic bite. A continuous root configuration has been forced to pay at least one unit at the two endpoints.
Nonreciprocity collapses the degree#
Boyd's exterior-orbit incidence bound says that if a reciprocal algebraic integer of degree \(n\) has \(v\) conjugates outside the unit circle and its Mahler product is nonreciprocal, then
$$ n<2v^2. $$
Here \(v=2\), so
$$ \deg P<8. $$
Reciprocity leaves only degrees four and six. The all-degree problem has become two equality checks. Very cute.
Degree four overshoots \(\varphi^2\)#
When \(\deg P=4\), the trace polynomial has degree two. Equation (3) gives
$$ M\ge\varphi^2, $$
where \(\varphi=(1+\sqrt5)/2\).
Equality forces \(A=B=\varphi\). If the two exterior trace roots have the same sign, \(Q\) repeats a root. If they have opposite signs, then
$$ Q(u)=u^2-5 $$
and
$$ x^2Q(x+x^{-1}) =x^4-3x^2+1 =(x^2-x-1)(x^2+x-1). $$
Either way irreducibility fails. Therefore
$$ \boxed{M(P)>\varphi^2} $$
in degree four.
Degree six cannot attain two#
When \(\deg P=6\), equation (3) gives
$$ M\ge2. $$
Equality forces the interior trace root to be \(0\) and \(A=B=\sqrt2\). Equal exterior signs again repeat a trace root. Opposite signs produce
$$ Q(u)=u\left(u^2-\frac92\right), $$
which is not in \(\mathbb Z[u]\).
So equality is impossible here too:
$$ \boxed{M(P)>2.} $$
Combining the two degrees proves (1).
What survives#
The surviving branch has reciprocal exterior product. This argument does not exclude it, and abstract overlap incidence cannot finish it: reciprocal products can sit at the fixed-point support patterns that the nonreciprocal degree bound deliberately avoids.
The next step has to use arithmetic or Galois realizability of the source polynomial. Counting exterior pairs is finished; the product's reciprocity is now the actual fault line.
Notebook references: S-0025, R-0562