Boubaker's Theorem/Proof of Uniqueness

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring.

Let $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring of $R$ whose zero is $0_D$ and whose unity is $1_D$.

Let $X \in R$ be transcendental over $D$.

Let $D \left[{X}\right]$ be the ring of polynomial forms in $X$ over $D$.

Consider the following properties:
 * $(1): \quad \displaystyle \sum_{k \mathop = 1}^N {p_n \left({0}\right)} = -2N$
 * $(2): \quad \displaystyle \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)} = 0$
 * $(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_n \left({x}\right)} {\mathrm d x}}\right|_{x \mathop = 0} = 0$
 * $(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_n^2 \left({x}\right)} {\mathrm d x^{2}}}\right|_{x \mathop = 0} = \frac 8 3 N \left({N^2-1}\right)$

where, for a given positive integer $n$, $p_n \in D \left[{X}\right]$ is a non-null polynomial such that $p_n$ has $N$ roots $\alpha_k$ in $F$.

Then the subsequence $\left \langle {B_{4n} \left({x}\right)}\right \rangle$ of the Boubaker polynomials is the unique polynomial sequence of $D \left[{X}\right]$ which verifies simultaneously the four properties $(1) - (4)$.

Proof
Let:
 * $\left({R, +, \circ}\right)$ be a commutative ring
 * $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring of $R$ whose zero is $0_D$ and whose unity is $1_D$
 * $X \in R$ be transcendental over $D$.

It has been demonstrated that the Boubaker Polynomials sub-sequence $B_{4n} \left({x}\right)$, defined in $D \left[{X}\right]$ as:
 * $\displaystyle B_{4n} \left({x}\right) = 4 \sum_{p \mathop = 0}^{2n} \frac {n-p} {4n-p} \binom {4n-p} p \left({-1}\right)^p x^{2 \left({2n - p}\right)}$

satisfies the properties:
 * $(1): \quad \displaystyle \sum_{k \mathop = 1}^N {p_n \left({0}\right)} = -2N$
 * $(2): \quad \displaystyle \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)} = 0$
 * $(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_n \left({x}\right)}{\mathrm d x}}\right|_{x \mathop = 0} = 0$
 * $(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_n^2 \left({x}\right)}{\mathrm d x^2}}\right|_{x \mathop = 0} = \frac 8 3 N (N^2-1)$

with $\left. {\alpha_k}\right |_{k = 1 \,.\,.\, N}$ roots of $ B_{4n}$.

Suppose there exists another $4n$-indexed polynomial $ q_{4n} \left({x}\right)$, with $N$ roots $\left.{\beta_k }\right|_{k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$.

Let:
 * $\displaystyle B_{4n} \left({x}\right) = \sum_{p \mathop = 0}^{2n} a_{4n, p} x^{2 \left({2n - p}\right)}$

and:
 * $\displaystyle q_{4n} \left({x}\right) = \sum_{p \mathop = 0}^{2n} b_{4n, p} x^{2 \left({2n - p}\right)}$

and:
 * $\displaystyle \mathrm d_{4n, p} = a_{4n, p} - b_{4n, p}$ for $p = 0 \,.\,.\, 2n$

then, simultaneous expressions of conditions $(1)$ and $(3)$ give:


 * $ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4n, 2n} = 0$
 * $ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4n, 2n-2} = 0$

It has also been demonstrated that $ B_{4n}$ has exactly $4n-2$ real roots inside the domain $\left[{-2 \,.\,.\, 2}\right]$.

So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n-3}$.

Finally, since $ B_{4n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n}$, with evident solution:
 * $\displaystyle \left.{\mathrm d_{4n, p}}\right|_{p \mathop = 0 \,.\,.\, 2n} = 0 $

and consequently:
 * $\displaystyle \left. { a_{4n, p}}\right|_{p \mathop = 0 \,.\,.\, 2n} = \left.{ b_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n}$

which means:
 * $q_{4n} \left({x}\right) = B_{4n} \left({x}\right) $