In both cases the proof is obtained by a contradiction with a hypothesis on the coefficients $\nu(N)$ (resp. $\nu^{\star}(N)$) appearing in some rational function of the form $\frac{\nu(2)}{x^{N-2}}+\cdots+\frac{\nu(N-2)}{x^{2}}+\nu(N)+O_{N}$ where $O_{N}$ is some polynomial function, for the Goldbach conjecture; and of the form $\frac{\nu^{\star}(4)}{x^{N-4}}+\cdots+\frac{\nu^{\star}(N-4)}{x^{4}}+\nu^{\star}(N)+O^{\star}{N}$ where $O^{\star}{N}$ is some polynomial function, for the twins prime numbers conjecture. In the first case, $\nu(N)$ represents the number of times where the even number $N$ is sum of two prime numbers, and one makes the assumption that there exist $N_{0}$ such that $\nu(N)\neq 0$ for $N\leq N_{0}$ and $\nu(N_{0}+2)=0$. In the second case, $\nu^{\star}(N)$ is equal to $2$ or $0$ according to whether $N$, multiple of $4$, is sum of two twins prime numbers or not, and one makes the assumption that there is $N_{0}$ such that $\nu^{\star}(N_{0})\neq 0$ and $\nu^{\star}(N)= 0\ \forall N\geq N_{0}+4$. Now, the two proofs are built with the same arguments with few differences, and we use the same linear algebra calculations.
