=== 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 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: :$\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$ which gives: :$B_{4 n} \left({0}\right) = \dfrac {n - 4 n / 2} {n - n / 2} \dbinom {n - n / 2} {n / 2} = -2$ and finally: : $(1): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({0}\right) = \sum_{k \mathop = 1}^N -2 = - 2 N$ ;Property $(2)$: We have, for given integer $n$, $ B_{4n} \in D \left[{X}\right]$ is a non-null polynomial with $N$ roots $\alpha_k$ in $F$.   Since: :$B_{4 n} \left({\alpha_k}\right) = 0$ then the equality: : $(2): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({\alpha_k}\right) = 0$ holds. ;Property $(3)$: According to the closed form of the the Boubaker Polynomials: :$\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$ We have: :$\displaystyle \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 1} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) x^{n - 2 p - 1}$           The minimal power in this expansion is obtained for $p = {\left\lfloor{n / 2}\right\rfloor - 1}$, hence: :$\dfrac {\mathrm d B_{4 n} } {\mathrm d x} \left({0}\right) = 0$ and the equality: : $(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} } \right\vert_{x \mathop = 0} = 0$ holds.   ;Property $(4)$: Starting from the closed form of the the Boubaker Polynomials: :$\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$ we have consequently: :$\displaystyle \frac {\mathrm d^2 B_{4n} \left({x}\right)} {\mathrm d x^2} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 2} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) x^{n - 2 p - 2}$           The minimal power in this expansion is obtained for $p = \left\lfloor{n / 2}\right\rfloor - 2$, hence: :$\displaystyle \frac {\mathrm d^2 B_{4 n} } {\mathrm d x^2} \left({0}\right) = \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) 0$ and the equality: : $(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d^2 B_{4 n} \left({x}\right)} {\mathrm d x^2}}\right\vert_{x \mathop = 0} = \frac 8 3 N \left({N^2 - 1}\right)$ holds. {{qed}} {{Proofread}} === Proof of Uniqueness === {{:Boubaker's Theorem/Proof of Uniqueness}} {{finish}} {{Namedfor|Boubaker Boubaker|cat = Boubaker}} Category:Boubaker Polynomials 3symgdip87gorka3pmkubxjyvqfnlc2