Boubaker's Theorem

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 $D$.

Finally, consider the following properties:
 * $(1): \quad \displaystyle \sum_{k=1}^N {p_n \left({0}\right)} = -2N$
 * $(2): \quad \displaystyle \sum_{k=1}^N {p_n \left({\alpha_k}\right)} = 0$
 * $(3): \quad \displaystyle \left.{\sum_{k=1}^N \frac {dp_{n}(x)}{dx}}\right|_{x = 0} = 0$
 * $(4): \quad \displaystyle \left.{\sum_{k=1}^N \frac {dp_{n}^{2}(x)}{dx^{2}}}\right|_{x = 0} = \frac 8 3 N \left({N^2-1}\right)$

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 of validity
We first prove that the Boubaker Polynomials sub-sequence $ B_{4n}(x)$, defined in $D \left[{X}\right]$ verifies properties $(1)$, $(2)$, $(3)$ and $(4)$.

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 $D$.

We have the closed form of the the Boubaker Polynomials:
 * Property $(1)$:
 * $\displaystyle B_n \left({x}\right) = \sum_{p=0}^{\lfloor n/2\rfloor} \frac {n-4p} {n-p} \binom {n-p} p \left({-1}\right)^p x^{n-2p}$

which gives:
 * $\displaystyle B_{4n} \left({0}\right) = \frac {n-4n/2} {n-n/2} \binom {n-n/2} { n/2 }=-2$

and finally:
 * $(1): \quad \displaystyle \sum_{k=1}^N{B_{4n}(0)}= \sum_{k=1}^N{-2}=-2N $

We have, for given integer $n$, $ B_{4n} \in D \left[{X}\right]$ is a non-null polynomial with $N$ roots $\alpha_k$ in $F$.
 * Property $(2)$:

Since:
 * $\displaystyle B_{4n} (\alpha_k)=0$

then the equality:
 * $(2): \quad \displaystyle \sum_{k=1}^N { B_{4n}(\alpha_k)}=0$

holds.

According to the closed form of the the Boubaker Polynomials:
 * Property $(3)$:
 * $\displaystyle B_n \left({x}\right) = \sum_{p=0}^{\lfloor n/2\rfloor} \frac {n-4p} {n-p} \binom {n-p} p \left({-1}\right)^p x^{n-2p}$

We have:
 * $\displaystyle \frac {d B_{4n} (x)}{dx}= \sum_{p=0}^{\lfloor n/2\rfloor -1} \frac {n-4p} {n-p} \binom {n-p} p \left({-1}\right)^p {(n-2p)}x^{n-2p-1}$

The minimal power in this expansion is obtained for $p={\lfloor n/2\rfloor-1}$, hence :
 * $\displaystyle \frac {dB_{4n}} {dx} (0)=0$

and the equality:
 * $(3): \quad \displaystyle \left.{\sum_{k=1}^N \frac {d B_{4n}(x)}{dx}}\right|_{x=0} = 0$

holds.

Starting from the closed form of the the Boubaker Polynomials:
 * Property $(4)$:
 * $\displaystyle B_n \left({x}\right) = \sum_{p=0}^{\lfloor n/2\rfloor} \frac {n-4p} {n-p} \binom {n-p} p \left({-1}\right)^p x^{n-2p}$

we have consequently:
 * $\displaystyle \frac {d{^2} B_{4n} (x)}{dx{^2}}= \sum_{p=0}^{\lfloor n/2\rfloor -2} \frac {n-4p} {n-p} \binom {n-p} p \left({-1}\right)^p {(n-2p)}{(n-2p-1)}x^{n-2p-2}$

The minimal power in this expansion is obtained for $p={\lfloor n/2\rfloor-2}$, hence :
 * $\displaystyle \frac{d^2 B_{4n}} {dx^2} (0) = \left({-1}\right)^p {n-2p}{n-2p-1}0$

and the equality:
 * $(4): \quad \displaystyle \left.{\sum_{k=1}^N \frac {d^2 B_{4n} (x)} {dx^2}}\right|_{x = 0} = \frac 8 3 N(N^2-1) $

holds.