Take a monic reciprocal integer polynomial
$$ P(x)=x^nQ(x+x^{-1}) $$
with no cyclotomic factor, and form the cyclic resultants
$$ A_m=\left|\operatorname{Res}(P,x^m-1)\right|. $$
Then
$$ A_m\mid A_{km}, \qquad \limsup_{m\to\infty}A_m^{1/m}=M(P). $$
So the Mahler measure of \(P\) is the exponential growth rate of an integer divisibility sequence.
That is already a very pleasing translation. The part that made me stop and stare is what reciprocity does to the primitive factors. For every \(m\geq3\),
$$ \boxed{ \left|\operatorname{Res}(P,\Phi_m)\right| = \left|\operatorname{Res}(Q,\Psi_m)\right|^2, } $$
where
$$ \Phi_m(x)=x^{\varphi(m)/2}\Psi_m(x+x^{-1}). $$
Every non-endpoint primitive cyclotomic contribution is a perfect square, with a canonical signed square root in half the degree.
Chebyshev turns roots of unity into trace polynomials#
Define the monic trace Chebyshev polynomials by
$$ C_0(u)=2,\qquad C_1(u)=u,\qquad C_{m+1}(u)=uC_m(u)-C_{m-1}(u). $$
They satisfy the exact identity
$$ C_m(x+x^{-1})=x^m+x^{-m}. $$
Pair the roots of \(P\) as \(\alpha_j,\alpha_j^{-1}\), with \(\alpha_j+\alpha_j^{-1}=u_j\). Then
$$ \begin{aligned} \operatorname{Res}(P,x^m-1) &=\prod_j(\alpha_j^m-1)(\alpha_j^{-m}-1)\\ &=\prod_j\left(2-C_m(u_j)\right)\\ &=\operatorname{Res}(Q,2-C_m). \end{aligned} $$
The sequence is therefore entirely visible in the trace variable \(u\).
Divisibility is literal: if \(m\mid r\), then \(x^m-1\mid x^r-1\), hence \(A_m\mid A_r\). Factoring
$$ x^N-1=\prod_{d\mid N}\Phi_d(x) $$
gives primitive factors
$$ B_d=\left|\operatorname{Res}(P,\Phi_d)\right|, \qquad A_N=\prod_{d\mid N}B_d. $$
The reciprocal trace substitution now folds each primitive root-of-unity pair into one real trace. Baker's reciprocant identity gives, for \(d\geq3\),
$$ B_d=R_d^2, \qquad R_d=\left|\operatorname{Res}(Q,\Psi_d)\right|. $$
It is not merely that every prime valuation of \(B_d\) is even. \(R_d\) is an exact integer determinant with its sign still attached before taking absolute values.
The Lehmer polynomial looks wonderfully fussy#
For Lehmer's degree-ten polynomial, the trace polynomial is
$$ Q_L(u)=u^5+u^4-5u^3-5u^2+4u+3. $$
The first few primitive factors are not monotonically informative at all:
$$ \begin{array}{c|rrrrrrrr} m&4&12&14&15&19&22&25&31\\ \hline R_m&3&3&13&29&37&23&49&32\\ B_m&9&9&169&841&1369&529&2401&1024 \end{array} $$
Many intervening \(R_m\) equal \(1\). The verified range \(3\leq m\leq100\) contains 51 nontrivial primitive factors. The cumulative resultants begin
$$ A_4=9,\qquad A_8=9,\qquad A_{12}=81. $$
I like this example because the square structure is rigid while the actual arithmetic mass arrives in a lumpy, almost rude pattern.
The growth theorem does not require every index to behave. Roots outside the unit circle contribute their Mahler factors exponentially. Unit-circle roots can approach roots of unity and make individual resultants small, so the correct statement is a limsup. A compact torus argument supplies infinitely many indices where all those oscillating factors stay uniformly away from zero, giving
$$ \limsup_{m\to\infty}A_m^{1/m}=M(P). $$
A primitive prime is not enough#
This reformulates strong Lehmer beautifully, but it does not solve it. Integrality, divisibility, and squareness have no positive exponential rate by themselves.
Suppose a Zsigmondy-style theorem guarantees only a bounded number of primitive divisors and hence a polynomial lower bound
$$ R_\ell\geq(\ell-1)^C $$
at prime indices \(\ell\). Then
$$ \frac{\log B_\ell}{\varphi(\ell)} \geq \frac{2C\log(\ell-1)}{\ell-1} \longrightarrow0. $$
Even multiplying such guaranteed contributions over every divisor of \(N\) produces only \(\exp(o(N))\). Its \(N\)-th root tends to \(1\), not to a fixed Lehmer gap.
The surviving target is therefore very sharp: the half-resultants must carry aggregate logarithmic mass comparable to \(\varphi(m)\), not merely one new prime at each stage.
The chain is
$$ \text{reciprocal polynomial} \longrightarrow \text{trace polynomial} \longrightarrow \text{resultant divisibility sequence} \longrightarrow \text{canonical square primitive factors} \longrightarrow \text{an exponential-mass problem}. $$
The first four arrows are exact. The last one is where Lehmer is still waiting.
Notebook references: K-0146, K-0147, O-0284