# Constructing a polynomial with arbitrary Mahler Measure

Let $p(z) \in \mathbb{C}[z]$, and $p(z) = a(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n).$

Here is a definition of Mahler measure.

$M: \mathbb{C}[z] \to \mathbb{R} \text{, with } M(p) = |a| \prod_{i=1}^n \max\{1, |\alpha_i|\}$

Lehmer conjectured,

Lehmer’s Mahler-measure Conjecture:
There exists a real number $\mu$ such that $\forall p(x) \in \mathbb{Z}[x], M(p) = 1$ or $1 < \mu \leq M(p)$. Lehmer suggested

$M(p_\mu) = \mu = 1.176280818...$

where $p_\mu$ is given as

$p_\mu = x^{10} - x^9 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 + 1$.

Now, let us “try” to construct a polynomial $p(x)$, such that $1 < M(p) < \mu$.

First, consider $1 < \alpha < \mu$. Hence, $(x-\alpha)(x-\frac{1}{\alpha}) = x^2 - (\alpha+\frac{1}{\alpha}) + 1$; however, $\alpha + \frac{1}{\alpha} \not \in \mathbb{Z}$, so we need to calibrate the polynomial with more terms.

Let

$p_N(x) = (x^2-(\alpha+\frac{1}{\alpha})x + 1) \prod_{i=1}^N (x- \beta_i)(x-\overline{\beta_i})$

where $\forall i, \beta_i, \overline{\beta_i} \in \mathbb{C}$ and $\beta_i \overline{\beta_i} = {|\beta_i|}^2 = 1$.

Thus, $(x-\beta_i)(x-\overline{\beta_i}) = x^2-2\Re(\beta_i)x+1$.

Let us substitute $(\alpha+\frac{1}{\alpha})$ by $b_0$ and $2\Re(\beta_i)$ by $b_i$; then, we may rewrite the polynomial $p(x)$ as

$p_{N}(x) = \prod_{i=0}^N (x^2-b_ix+1)$

Here are the first few cases;

$n = 0$, then $p_0(x) = x^2-b_0x+1$

$n = 1$, then $p_1(x) = x^4 - (b_0+b_1) x^3 + (b_0b_1+2)x^2 - (b_0+b_1)x + 1$

$n = 2$, then $p_2(x) = x^6 - (b_0+b_1+b_2)x^5 + (b_0b_1+b_0b_2+b_1b_2+3)x^4-(2b_0+2b_1+2b_2+b_0b_1b_2)x^3 + (b_0b_1+b_0b_2+b_1b_2+3)x^2 - (b_0+b_1+b_2)x + 1$

Obviously, for given $n$, $x^k$ and $x^{2n-k}$ share the same coefficient.

Since the polynomial is getting messier, we need a new notation; let $\pi_k^n$ denote the sum of product of $k$-combination from set of $\{b_i\}_{i=0}^{n-1}$. For instance, $\pi_2^3 = b_0b_1 + b_0b_2 + b_1b_2$.

Now, in stead of writing down the whole family of polynomials into a table. Let us define a matrix $X$ where each entry $x_{k,j}$, (the $k$-th row and $j$-th column) represents the coefficient of $x^{j-1}$ from $p_{k-1}(x)$. The entry is left blank if the coefficient is 0, just to avoid confusion.

The coefficients seem pretty random though, there is a specific pattern lurking.

Consider $k \leq 2i$. Then, $x^k$ term in $p_i(x)$ is only determined by $x^{k-2}, x^{k-1}$ and $x^k$ terms from $p_{i-1}(x)$.

For example, consider $x_{5,5}$; it is $10 + 3\pi_2^5 + \pi_4^5$. According to the idea presented above, this term is created by

$(4+\pi_2^4)x^2 \cdot x^2 + (-3\pi_1^4 - \pi_3^4)x^3 \cdot (-b_4) x + (6 + 2\pi_2^4 + \pi_4^4)x^4 \cdot 1$

Since the degree of $x$ is all 4, let us not consider about the $x^4$ for now. Then,

$4 + 6 = 10$

$\pi_2^4 + (-b_4)(-3\pi_1^4) + 2\pi_2^4 = 3\pi_2^4 + 3b_4\pi_1^3 = 3 \pi_2^5$

and

$(-b_4)(-\pi_3^4) + \pi_4^4 = \pi_4^5$.

Thus, it nicely transforms to the term, as desired.

With this nice property, I induced a following equation.

Lee Equation:

As we move back to the matrix, let us assume that $p_{n-1}(x)$ has Mahler measure less than $\mu$. Then, every term along the $n$-th row is composed of integers; obviously it leads to the conclusion that for all $k \leq n, \pi_k^n$ are integers.

If Lehmer’s conjecture turns out to be true, then it implies for every $n$, there exists no sets of $\{b_i\}_{i=0}^{n-1}$ that renders $\pi_k^n \in \mathbb{Z}$ for every $k$.