Theorem: Boubaker's Theorem/Proof of Uniqueness

Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Consider the following properties:
{{begin-eqn}}
{{eqn | n = 1
      | l = \sum_{k \mathop = 1}^N {\map {p_n} 0}
      | r = -2N
}}
{{eqn | n = 2
      | l = \sum_{k \mathop = 1}^N {\map {p_n} {\alpha_k} }
      | r = 0
}}
{{eqn | n = 3
      | l = \valueat {\sum_{k \mathop = 1}^N \frac {\d \map {p_x} x} {\d x} } {x \mathop = 0}
      | r = 0
}}
{{eqn | n = 4
      | l = \valueat {\sum_{k \mathop = 1}^N \frac {\d \map { {p_n}^2} x} {\d x^2} } {x \mathop = 0}
      | r = \frac 8 3 N \paren {N^2 - 1}
}}
{{end-eqn}}
where, for a given positive integer $n$, $p_n \in D \sqbrk X$ is a non-null polynomial such that $p_n$ has $N$ roots $\alpha_k$ in $F$.
Then the subsequence $\sequence {\map {B_{4 n} } x}$ of the Boubaker polynomials is the unique polynomial sequence of $D \sqbrk X$ which verifies simultaneously the four properties $(1) - (4)$.

Proof:
Let:
:$\struct {R, +, \circ}$ be a commutative ring
:$\struct {D, +, \circ}$ be an integral subdomain 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 $\map {B_{4 n} } x$, defined in $D \sqbrk X$ as:
:$\ds \map {B_{4 n} } x = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \paren {-1}^p x^{2 \paren {2 n - p} }$
satisfies the properties:
{{begin-eqn}}
{{eqn | n = 1
      | l = \sum_{k \mathop = 1}^N {\map {p_n} 0}
      | r = -2N
}}
{{eqn | n = 2
      | l = \sum_{k \mathop = 1}^N {\map {p_n} {\alpha_k} }
      | r = 0
}}
{{eqn | n = 3
      | l = \valueat {\sum_{k \mathop = 1}^N \frac {\d \map {p_x} x} {\d x} } {x \mathop = 0}
      | r = 0
}}
{{eqn | n = 4
      | l = \valueat {\sum_{k \mathop = 1}^N \frac {\d \map { {p_n}^2} x} {\d x^2} } {x \mathop = 0}
      | r = \frac 8 3 N \paren {N^2 - 1}
}}
{{end-eqn}}
with $\valueat {\alpha_k} {k \mathop = 1 \,.\,.\, N}$ roots of $B_{4 n}$.
Suppose there exists another $4 n$-indexed polynomial $\map {q_{4 n} } x$, with $N$ roots $\valueat {\beta_k} {k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$.
Let:
:$\ds \map {B_{4 n} } x = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \paren {2 n - p} }$
and:
:$\ds \map {q_{4 n} } x = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \paren {2 n - p} }$
and:
:$\d_{4 n, p} = a_{4 n, p} - b_{4 n, p}$ for $p = 0 \,.\,.\, 2 n$
then, simultaneous expressions of conditions $(1)$ and $(3)$ give:
:$\quad \ds \sum_{k \mathop = 1}^N \d_{4 n, 2 n} = 0$
:$\quad \ds \sum_{k \mathop = 1}^N \d_{4 n, 2 n - 2} = 0$
It has also been demonstrated that $B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\closedint {-2} 2$.
So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n - 3}$.
Finally, since $B_{4 n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n}$, with evident solution:
:$\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n} = 0$
and consequently:
:$\bigvalueat {a_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n} = \bigvalueat {b_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n}$
which means:
:$\map {q_{4 n} } x = \map {B_{4 n} } x$
{{qed}}