Theorem: Boubaker's Theorem

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 $D$.
Finally, 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 \map {\frac {\d {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:
=== Proof of validity ===
We first prove that the Boubaker Polynomials subsequence $\sequence {\map {B_{4 n} } x}$, defined in $D \sqbrk X$ verifies properties $(1)$, $(2)$, $(3)$ and $(4)$.
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 $D$.
;Property $(1)$:
We have the closed form of the the Boubaker Polynomials:
:$\ds \map {B_n} x = \sum_{p \mathop = 0}^{\floor {n / 2} } \frac {n - 4 p} {n - p} \binom {n - p} p \paren {-1}^p x^{n - 2 p}$
which gives:
:$\map {B_{4 n} } 0 = \dfrac {n - 4 n / 2} {n - n / 2} \dbinom {n - n / 2} {n / 2} = -2$
and finally:
:$(1): \quad \ds \sum_{k \mathop = 1}^N \map {B_{4 n} } 0 = \sum_{k \mathop = 1}^N -2 = -2 N$
;Property $(2)$:
We have, for given integer $n$, $ B_{4n} \in D \sqbrk X$ is a non-null polynomial with $N$ roots $\alpha_k$ in $F$.  
Since:
:$\map {B_{4 n} } {\alpha_k} = 0$
then the equality:
:$(2): \quad \ds \sum_{k \mathop = 1}^N \map {B_{4 n} } {\alpha_k} = 0$
holds.
;Property $(3)$:
According to the closed form of the the Boubaker Polynomials:
:$\ds \map {B_n} x = \sum_{p \mathop = 0}^{\floor {n / 2} } \frac {n - 4 p} {n - p} \binom {n - p} p \paren {-1}^p x^{n - 2 p}$
We have:
:$\ds \frac {\d \map {B_{4 n} } x} {\d x} = \sum_{p \mathop = 0}^{\floor {n / 2} - 1} \frac {n - 4 p} {n - p} \binom {n - p} p \paren {-1}^p \paren {n - 2 p} x^{n - 2 p - 1}$  
The minimal power in this expansion is obtained for $p = {\floor {n / 2} - 1}$, hence:
:$\map {\dfrac {\d B_{4 n} } {\d x} } 0 = 0$
and the equality:
:$(3): \quad \ds \valueat {\sum_{k \mathop = 1}^N \frac {\d \map {B_{4 n} } x} {\d x} } {x \mathop = 0} = 0$
holds.
;Property $(4)$:
Starting from the closed form of the the Boubaker Polynomials:
:$\ds \map {B_n} x = \sum_{p \mathop = 0}^{\floor {n / 2} } \frac {n - 4 p} {n - p} \binom {n - p} p \paren {-1}^p x^{n - 2 p}$
we have consequently:
:$\ds \frac {\d^2 \map {B_{4 n} } x} {\d x^2} = \sum_{p \mathop = 0}^{\floor {n / 2} - 2} \frac {n - 4 p} {n - p} \binom {n - p} p \paren {-1}^p \paren {n - 2 p} \paren {n - 2 p - 1} x^{n - 2 p - 2}$      
The minimal power in this expansion is obtained for $p = \floor {n / 2} - 2$, hence:
:$\map {\dfrac {\d^2 B_{4 n} } {\d x^2} } 0 = \paren {-1}^p \paren {n - 2 p} \paren {n - 2 p - 1}0$
and the equality:
:$(4): \quad \ds \valueat {\sum_{k \mathop = 1}^N \frac {\d^2 \map {B_{4 n} } x} {\d x^2}} {x \mathop = 0} = \frac 8 3 N \paren {N^2 - 1}$
holds.
{{qed}}
{{Proofread}}