I don't have any new content to add, but I did notice some context:

A "lattice" in a real vector space $V$ is the ${\mathbb Z}$-span of an
${\mathbb R}$-basis of $V$. Any ordered ${\mathbb R}$-basis whose ${\mathbb Z}$-span is the lattice $L$ is said to be an "ordered base" of $L$.

Let $L$ be a lattice in a real inner product space $V$. (The inner product being, by definition, symmetric, bilinear and positive definite.) The "dual lattice" to $L$ in $V$ consists of vectors $v\in V$ such that, for all $w\in L$, we have:
$\langle v,w\rangle\in{\mathbb Z}$. The dual of $L$ will be denoted $L^*$.

Let $B$ be an ordered basis in a real inner product space $V$. Then the "Gramian matrix" of $B$ is the matrix whose $i,j$-entry is $\langle B_i,B_j\rangle$.

A matrix is "integral" if all of its entries are integers.

Lemma: Let $B$ be an ordered base of a lattice $L$ in a real inner product space. Then: $L^*\subseteq L$ iff the inverse of the Gramian matrix of $B$ is integral.

The proof of this lemma is not hard.

Now fix a positive integer $d$ and let $V$ be the real inner product space consisting of real polynomials ${\mathbb R}\to{\mathbb R}$ of degree $<d$, with inner product given by:
$\langle P,Q\rangle=\int_0^1 PQ$.

Let $M$ be the lattice in $V$ consisting of all polynomials in $V$ that have integer coefficients.

Let ${\bf1}$ denote the constant function
${\mathbb R}\to{\mathbb R}$ with value $1$.
Let ${\bf x}:{\mathbb R}\to{\mathbb R}$ denote the identity function.
Let $B:=({\bf1},{\bf x},{\bf x}^2,...,{\bf x}^{d-1})$, an ordered base of $M$.

The Gramian matrix of $B$ is the Hilbert $d\times d$ matrix, so, by the lemma, we wish to show that $M^*\subseteq M$.

Let $A:=(\tilde P_0,...\tilde P_{d-1})$, the ordered list of the $0$th through $(d-1)$st shifted Legendre polynomials. This is an ordered basis of $V$. Let $L$ be the ${\mathbb Z}$-span of $A$. The $L$ is a lattice in $V$.

Since the shifted Legendre polynomials have integer coefficients, $L\subseteq M$. It follows that $L^*\supseteq M^*$.

The Gramian matrix of $A$ is diagonal, with each diagonal entry in the set
$\{1,1/2,1/3,1/4,\dots\}$. So, by the lemma, $L^*\subseteq L$.

Then $M^*\subseteq L^*\subseteq L\subseteq M$. QED

The Art of Computer Programming. $\endgroup$1more comment