Lehmer’s Mahler Measure Problem

Lehmer’s Mahler Measure Problem

Definitions and mathematical statement

If a monic polynomial $P\in{\mathbb{Z}}[X]$ factors as $P(X)=\prod_{j=1}^d(X-\alpha_j)$ over ${\mathbb{C}}$, its multiplicative Mahler measure is \[ M(P)=\prod_{j=1}^d\max\{1,|\alpha_j|\}. \] The selected gap assertion is that an absolute $\varepsilon_0>0$ exists with $M(P)>1\Longrightarrow M(P)\ge1+\varepsilon_0$ for every monic integer polynomial of arbitrary degree. Equivalently, either establish a gap immediately above one or produce a sequence with $M(P_j)>1$ and $M(P_j)\to1$. No fixed-degree bound can substitute for an absolute bound across all degrees, and the polynomials need not be irreducible.

Formulation and concept references: [S1].

Short English statement

Is there a universal gap above one among Mahler measures of monic integer polynomials, or can such measures approach one from above?

Research attempt

Current frontier.

For unrestricted degree the established general lower bounds have Dobrowolski-type dependence on the degree, while Smyth's theorem gives a uniform gap for nonreciprocal algebraic integers of measure greater than one.[E1] The Schinzel--Zassenhaus theorem gives a lower bound for the house $\max_j|\alpha_j|$, not the required uniform lower bound for the product $M(P)$. Dimitrov's Theorem 1 gives $\max_j|\alpha_j|\ge2^{1/(4d)}$ for monic irreducible noncyclotomic polynomials other than $X$; the Annals publication page records acceptance in August 2026.[E2] For each fixed bound on the number of nonzero monomials, there is a gap above measure one, independent of the exponents; that bound cannot be held fixed in a hypothetical counterexample sequence.[E3] None of these statements substitutes for the selected degree-independent assertion over all monic integer polynomials.

Attack route.

Resultant methods would need arithmetic amplification that does not deteriorate with degree. A sparse-polynomial construction cannot work with bounded support size, by the preceding theorem.[E3] The route developed here combines exact integer power sums with positive semidefinite moment matrices of points on the unit circle. Reciprocal pairing causes a quadratic, rather than linear, radial error. This produces explicit lower-bound certificates and a normalization that survives the substitution $X\mapsto X^N$. The unresolved question is whether certificates of uniformly positive normalized strength always exist, not merely whether some moment eventually detects an off-circle root.

Attempt.

1. Reduction and normalization. Write $m(P)=\log M(P)$. Multiplicativity of $M$ follows immediately by concatenating roots. A monic integer polynomial with $M(P)>1$ therefore has a monic irreducible factor $Q$ with $1<M(Q)\le M(P)$. Factors $X$ can be removed. If $|Q(0)|\ge2$, then $M(Q)\ge|Q(0)|\ge2$. The unresolved small-measure regime is thus algebraic units. Smyth's nonreciprocal bound further reduces a prospective sequence tending to one to reciprocal units.[E1] For an irreducible reciprocal unit of measure greater than one we can use $Q(X)=X^dQ(1/X)$ and $Q(0)=1$: the anti-reciprocal possibility would have a root at $1$ and hence could not be such an irreducible polynomial. The estimates below also apply to reducible reciprocal monic polynomials with constant term one.

Fix such a polynomial $P$, with roots counted with multiplicity, and set \[ \alpha_j=r_je^{i\theta_j},\quad m=\sum_{r_j>1}\log r_j,\quad h=\max_{r_j>1}\log r_j>0, \qquad s_k=\sum_{j=1}^d\alpha_j^k\in{\mathbb{Z}}\quad(k\ge0). \] Put $s_0=d$ and extend by $s_{-k}=s_k$. Reciprocal symmetry justifies this extension as the actual negative power sum; reality of the coefficients makes these sums real. Define the projected phase moments \[ u_k=\sum_{j=1}^de^{ik\theta_j},\qquad u_{-k}=u_k. \] The phases occur in conjugate pairs, so $u_k$ is real.

Lemma 206.1 (quadratic reciprocal radial error). For every integer $k$, \[ |s_k-u_k|\le k^2 h m\cosh(|k|h). \tag{206.1} \]

Proof. An outside root $\alpha=e^{\ell+i\theta}$ is paired with the inside root $1/\overline\alpha=e^{-\ell+i\theta}$; the polynomial is both real and reciprocal, so this pairing respects multiplicities. Their contribution to $s_k-u_k$ is \[ \left(e^{k\ell}+e^{-k\ell}-2\right)e^{ik\theta} =2(\cosh(k\ell)-1)e^{ik\theta}. \] Unit-circle roots contribute zero. Since $2(\cosh t-1)\le t^2\cosh t$ for real $t$, summing over the outside roots gives \[ |s_k-u_k| \le k^2\cosh(|k|h)\sum_{r_j>1}(\log r_j)^2 \le k^2hm\cosh(|k|h). \] The elementary hyperbolic-cosine inequality follows by integrating $\cosh$ twice from zero, or termwise from its power series. All sums over roots are finite.$\square$

The quadratic cancellation would be lost by estimating each root and its reciprocal separately. Importantly $h$, the largest individual radial deviation, can be much smaller than the total deviation $m$. Replacing it by $m$ prematurely destroys useful substitution invariance.

2. Exact integer certificates from moment positivity. Let $a_0,\ldots,a_L$ be distinct nonnegative integer indices and $v=(v_0,\ldots,v_L)\in{\mathbb{Z}}^{L+1}$. Define \[ T=(s_{a_i-a_j})_{0\le i,j\le L},\qquad G=(u_{a_i-a_j})_{0\le i,j\le L}. \] The second matrix is positive semidefinite because \[ v^T Gv=\sum_{j=1}^d \left|\sum_{i=0}^L v_i e^{ia_i\theta_j}\right|^2\ge0. \] If the exact integer calculation gives $v^TTv=-q<0$, Lemma 206.1 therefore yields \[ q\le hm\,H_h(a,v),\qquad H_h(a,v)=\sum_{i,j=0}^L|v_iv_j|(a_i-a_j)^2 \cosh(|a_i-a_j|h). \] In particular, \[ \boxed{\displaystyle m(P)\ge\frac{q}{h\,H_h(a,v)}.} \tag{206.2} \] The denominator is positive for a negative certificate, since otherwise $v$ has at most one nonzero entry and $v^TTv=d\|v\|_2^2\ge0$. If a rigorous upper bound $\bar h\ge h$ is available, the weaker but certified estimate $m(P)\ge q/(\bar h H_{\bar h}(a,v))$ follows by monotonicity. Approximate eigenvectors may suggest $v$, but the sign of $v^TTv$ must be checked over the integers.

There is always some negative certificate when $m(P)>0$. Otherwise every $2\times2$ matrix with indices $0,k$ is positive semidefinite, giving $|s_k|\le d$ for all $k$. The power series \[ \sum_{k\ge0}s_kz^k=\sum_{j=1}^d\frac{1}{1-\alpha_jz} \tag{206.3} \] would then be analytic for $|z|<1$. An outside root produces a pole at $z=1/\alpha_j$ in that disk. Repeated copies of this root add identical nonzero residues; distinct roots have different pole locations, so the pole cannot cancel. This contradicts $206.3$, initially valid near zero and hence throughout its domain by analytic continuation. Thus some $|s_k|>d$, and $v=(1,-\operatorname{sgn}s_k)$ at indices $0,k$ is negative. This proves existence but gives no uniform strength for $206.2$.

3. A substitution stress test, and why the normalization helps. For any monic polynomial $Q$ and positive integer $N$, \[ M(Q(X^N))=M(Q),\qquad \operatorname{house}(Q(X^N))= \operatorname{house}(Q)^{1/N}. \tag{206.4} \] Indeed the $N$ roots above a nonzero root $\beta$ all have modulus $|\beta|^{1/N}$, and their factors in the product defining $M$ multiply to $\max(1,|\beta|)$. Zeros also obey the formula. Consequently small house alone cannot manufacture a Lehmer counterexample.

Moreover the power sums of $Q(X^N)$ are zero when $N$ does not divide $k$, and equal $N s_{k/N}(Q)$ when it does. Thus any fixed finite prefix of positive-index traces can be identically zero for a polynomial of measure greater than one. The cyclotomic polynomial product $X^{N\deg Q}+1$ has the same degree and the same prefix. A test using only a fixed initial trace window therefore cannot decide the original problem. Substitution may make a polynomial reducible, which is allowed by the original target; this example does not by itself refute every possible irreducibility-sensitive test.

Nevertheless $206.2$ passes a more demanding check. For reciprocal $Q$, replace $a_i$ by $Na_i$ and keep $v$ when passing to $Q(X^N)$. Then \[ q_N=Nq,\qquad h_N=h/N,\qquad H_{h_N}(Na,v)=N^2H_h(a,v). \] Hence the certificate value $q_N/(h_NH_{h_N}(Na,v))$ is exactly the original value. This is why keeping the quadratic error and the separate quantity $h$ is preferable to a crude bound involving only $km$. The method does not automatically lose its estimate merely because all exponents have been multiplied by $N$.

4. A test certificate with exact verification. Take the classical degree-ten test polynomial \[ P_\mathrm L(X)=X^{10}+X^9-X^7-X^6-X^5-X^4-X^3+X+1. \] Its role as Lehmer's example is historical; the calculations that follow are performed directly.[E1] Newton's identities give, without root approximations, \[ (s_0,\ldots,s_8)=(10,-1,1,2,1,4,4,6,1). \] With consecutive indices $a_i=i$, choose \[ v=(2,3,2,1,0,-1,-2,-3,-2). \] Exact integer multiplication gives \[ v^TTv=-106,\qquad \sum_{i,j=0}^8|v_iv_j|(i-j)^2=4352. \tag{206.5} \] The vector was found by rounding a numerically computed negative eigenvector. Only the integer identities $206.5$, not the floating-point eigenvalue, enter the certificate.

For a rigorous radial bound, use \[ P_\mathrm L(X)=X^5 Q(X+X^{-1}),\qquad Q(t)=t^5+t^4-5t^3-5t^2+4t+3. \] A Sturm sequence, with each member independently scaled by a positive constant, is \[ \begin{gathered} Q(t),\quad 5t^4+4t^3-15t^2-10t+4,\\ 54t^3+60t^2-90t-71,\quad96t^2+117t-38, \quad1167t+1390,\quad1. \end{gathered} \] Put $a=117628/100000$, $b=117629/100000$ and $t_a=a+a^{-1}$, $t_b=b+b^{-1}$. Direct rational sign evaluation gives \[ \begin{array}{c|rrrrrr} t&-\infty&-2&2&t_a&t_b&+\infty\\\hline \text{sign variations}&5&5&1&1&0&0. \end{array} \] Thus four roots of $Q$ lie in $(-2,2)$ and its remaining root lies in $(t_a,t_b)\subset(2,\infty)$. The corresponding roots of $P_\mathrm L$ are eight unit-circle roots and a positive pair $\lambda,\lambda^{-1}$ with $a<\lambda<b$. Since \[ e^{1/6}>1+\frac16+\frac1{72}+\frac1{1296}>b, \] we obtain $h=m(P_\mathrm L)=\log\lambda<1/6$. Finally $\cosh(4/3)<21/10$: sum its terms through degree ten and bound the tail beginning with $(4/3)^{12}/12!$ by the geometric ratio $(4/3)^2/(14\cdot13)$; the resulting rational upper bound is $6491958743731/3200164983675<21/10$. Equations $206.2$--$206.5$ now prove the explicit, albeit weak, bound \[ m(P_\mathrm L)>\frac{106}{(1/6)(21/10)4352} =\frac{265}{3808}. \tag{206.6} \] This is a check of the certificate mechanism on a known example, not an improved known lower bound for that polynomial or for the problem. Numerical evaluation gives $\lambda\approx1.1762808182599175$ and $m\approx0.1623576120077381$; using the actual $h$ in $206.2$ gives a certificate value approximately $0.09753$. Those decimals are numerical illustrations only; $206.6$ uses rational and exact integer verification.

5. An infinite construction that fails for an exact reason. The candidate family $X^{2n}+aX^n+1$, $a\in{\mathbb{Z}}$, has \[ M(X^{2n}+aX^n+1)= \begin{cases} 1,&|a|\le2,\\[2pt] (|a|+\sqrt{a^2-4})/2,&|a|\ge3. \end{cases} \] Solve $Y^2+aY+1=0$ and use $206.4$. Thus taking $n\to\infty$ can make every root close to the unit circle without decreasing the Mahler measure at all. The smallest measure greater than one in this entire family is $(3+\sqrt5)/2$. More generally, any hypothetical sequence with $M\downarrow1$ must have unbounded monomial count, by the established bounded-support theorem.[E3]

Stress test.

Reciprocity, including its interaction with complex conjugation, is essential to $206.1$; for a general unit the linear radial error need not cancel. The phase moment matrix is positive semidefinite for an explicit reason, whereas the integer trace matrix need not be. The Sturm calculation avoids assuming from a plot that all the other roots lie on the unit circle. Existence of some negative certificate follows from a pole argument, not from extrapolating the finite eigenvalue search.

The unresolved uniform step is precise. It would suffice to find an absolute $c>0$ such that every remaining irreducible reciprocal $P$ with $M(P)>1$ admits integer indices and a vector with \[ -v^TTv\ \ge\ c\,h H_h(a,v)>0. \tag{206.7} \] Then $206.2$, together with the established reductions, would give the requested absolute gap. No bound of the form $206.7$ has been proved. An integer negative quadratic form is at most $-1$, but its denominator in $206.2$ can grow; integrality alone cannot be used to erase that growth. A proof of $206.7$ would settle Lehmer's problem and so cannot be regarded as an innocuous numerical conditioning assertion.

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Outcome.

Outcome: Exploratory approach with unresolved gap.

The attempt derives a quadratic reciprocal-root estimate and an exact integer moment-certificate inequality, proves that some negative certificate always exists off the unit circle, and checks the method with an independently verifiable degree-ten certificate. Its normalized bound is invariant under power substitution, removing one concrete failure mode of house-based arguments. No claim is made that these notebook lemmas improve the established general frontier or are new in the literature.

A complete proof requires uniformly strong certificates such as $206.7$, or another argument preventing their strength from tending to zero along increasing-degree, increasing-support polynomials. The explicit families tested here do not yield a counterexample sequence, and the finite computation does not exclude such sequences in general.

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Sources

• S1 Lehmer's Mahler Measure Problem (MathWorld). https://mathworld.wolfram.com/LehmersMahlerMeasureProblem.html. Catalogue formulation source.
• E1 C. Smyth, The Mahler measure of algebraic numbers: a survey, in Number Theory and Polynomials, London Mathematical Society Lecture Note Series 352, Cambridge University Press $2008$, 322--349; arXiv:math/0701397v2. Sections 1, 4.2 and 5 discuss Lehmer's polynomial, Dobrowolski's bound and the nonreciprocal theorem. https://arxiv.org/abs/math/0701397.
• E2 V. Dimitrov, A proof of the Schinzel--Zassenhaus conjecture on polynomials, arXiv:1912.12545, Theorem 1; accepted version titled The Schinzel--Zassenhaus conjecture, Annals of Mathematics, forthcoming. The journal page records acceptance on 27 August 2026; consulted 27 September 2026. This is a house theorem, not a solution of the Mahler-measure gap problem. https://arxiv.org/abs/1912.12545. https://annals.math.princeton.edu/articles/22975.
• E3 E. Dobrowolski and C. Smyth, Mahler measures of polynomials that are sums of a bounded number of monomials, arXiv:1606.04376 $2016$, Theorem 2 and its proof in Section 4: isolation of zero in the set of logarithmic measures for bounded monomial count. https://arxiv.org/abs/1606.04376.