Someone could help me find some error in the reasoning:
We know, that the canonical decomposition of $n!$ is:
$\displaystyle n!=\prod_{p_{i}\leq n}p_{i}^{\alpha_{i}(n)}$, where:
$\displaystyle \alpha_{i}(n)=\sum_{t=1}^{r}\left \lfloor \frac{n}{p_{i}^{t}}\right \rfloor$.
It's possible use the canonical decomposition of $n!$ to found the canonical decomposition of $n$, let's see:
$$n=\frac{n!}{(n-1)!}=\frac{\displaystyle \prod_{p_{i}\leq n}p_{i}^{\alpha_{i}(n)}}{\displaystyle \prod_{p_{i}\leq n-1}p_{i}^{\alpha_{i}(n-1)}}$$
$\displaystyle n=\prod_{p_{i}\leq n}p_{i}^{\alpha_{i}(n)-\alpha_{i}(n-1)}=\prod_{p_{i}\leq n}p_{i}^{\beta_{i}(n)}$, where:
$$\beta_{i}(n)=\alpha_{i}(n)-\alpha_{i}(n-1)$$
Note an important propertie of $\beta_{i}(n)$:
$\beta_{i}(p_{j})=\delta_{i,j}$, this is because the unicity of the canonical decomposition:
$$n=p_{j}=\prod_{p_{i}\leq p_{j}}p_{i}^{\beta_{i}(p_{j})}$$
Let's use this resoults to probe the followin hypothesis:
Hypothesis
$\forall N > 1, \exists k \in \{0,1,2,...,N-1\}$, such that, the prime factorization of $N^{2}-k^{2}$ always has at least one solution in the form $p_{1}p_{2}$, $(N^{2}-k^{2}=p_{1}p_{2})$, where $p_{1}$ and $p_{2}$ are prime numbers and $p_1 \leq p_2$.
we know that:
$$(N-k)=\prod_{p_{i}\leq N-k}p_{i}^{\beta_{i}(N-k)}$$
$$(N+k)=\prod_{p_{i}\leq N+k}p_{i}^{\beta_{i}(N+k)}$$
then:
$$(N-k)(N+k)=\prod_{p_{i}\leq N+k}p_{i}^{\beta_{i}(N-k)+\beta_{i}(N+k)}$$
$\displaystyle N^{2}-k^{2}=\prod_{p_{i}\leq N+k}p_{i}^{\gamma_{i}(N,k)}=p_1 p_2$, where:
$\gamma_{i}(N,k)=\beta_{i}(N-k)+\beta_{i}(N+k)$, then:
$\gamma_{i}(N,k)=\alpha_{i}(N-k)-\alpha_{i}(N-k-1)+\alpha_{i}(N+k)-\alpha_{i}(N+k-1)$
This is where I need help to find some error in the reasoning:
Then, due to the uniqueness of prime factorization, we have: $\gamma_{1}(N,k)=1; \gamma_{2}(N,k)=1; \gamma_{s}(N,k)=0$.
We can use the equations $\gamma_{1}(N,k)=1$, $\gamma_{2}(N,k)=1$, and $\gamma_{s}(N,k)=0$ other cases, to prove the hypothesis.
To prove that there exists a value of $k$ that satisfies the hypothesis, we can consider the sum of the first two equations:
$\gamma_{1}(N,k)+\gamma_{2}(N,k)=2$
We know that $\gamma_{1}(N,k)$ and $\gamma_{2}(N,k)$ are both either equal to $1$ or equal to $0$, so their sum can only be $0$, $1$, or $2$. But since $\gamma_{1}(N,k) + \gamma_{2}(N,k) = 2$, the sum of their values is always less than or equal to $2$, $\gamma_{1}(N,k) + \gamma_{2}(N,k) \leq 2$.
This implies that there exist two distinct prime factors $p_1$ and $p_2$ such that $N^{2}-k^{2}=p_{1}p_{2}$ and $p_1 \leq p_2$, which satisfies the hypothesis.
To prove that such a $k$ always exists, we can consider that $\gamma_{s}(N,k)=0$, which implies that all prime factors of $N^{2}-k^{2}$ are distinct from $p_{1}$ and $p_{2}$. Since the product of two distinct prime factors is always greater than twice the smaller factor, we can ensure that there always exists a value of $k$ such that $p_1 \leq k < N$ and $p_1p_2=N^2-k^2 \geq 2k+1$ (since $p_1$ and $p_2$ are primes and therefore greater than or equal to $2$). This is a is a consequence of Bertrand's postulate, which states that there is always a prime number between $n$ and $2n$ for any integer $n > 1$. This property ensures that there are always enough prime factors available to satisfy the conditions required for the existence of a value of $k$ such that $p_1 \leq k < N$ and $p_1p_2=N^2-k^2 \geq 2k+1$.
Thus, the existence of a value of $k$ that satisfies the hypothesis is proven.
As we have shown earlier, for any positive integer $N$ greater than $1$, there exists a $k$ in the range from $0$ to $N-1$ such that $N^2 - k^2$ is the product of two primes, where one is less than or equal to the other. Therefore, for that value of $k$, we can write:
$$N^2−k^2=(N−k)(N+k)=p_lp_m$$
Where $p_l$ and $p_m$ are the two primes that are factors of $N^2 - k^2$. Furthermore, since one factor is greater than the other, we can without loss of generality take $p_l \leq p_m$. Therefore, the statement is true for every $N>1$.
Therefore for every $N>1$, there exists a $k$, $0\leq k\leq N-1$, such that $N-k=p_l$ and $N+k=p_m$, where $p_l\leq p_m$, $p_l$ and $p_m$ are primes. This is because $p_l$ and $p_m$ are factors of $(N-k)(N+k)$ and moreover, $p_l \leq N-k < N < N+k \leq p_m$. Therefore, $p_l$ and $p_m$ must be equal to $N-k$ and $N+k$ (in some order) since they are the only prime numbers between $N-k$ and $N+k$. Note that: $$\gamma_{i}(N,k)=\alpha_{i}(N-k)-\alpha_{i}(N-k-1)+\alpha_{i}(N+k)-\alpha_{i}(N+k-1)$$
$$\gamma_{i}(N,k)=\alpha_{i}(p_{l})-\alpha_{i}(p_{l}-1)+\alpha_{i}(p_{m})-\alpha_{i}(p_{m}-1)$$
$$\gamma_{i}(N,k)=\beta_{i}(p_{l})+\beta_{i}(p_{m})$$
then:
$$\gamma_{i}(N,k)=\delta_{i,l}+\delta_{i,m}=\begin{cases} 1 &\text{if $i=l$}\\ 1 & \text{if $i=m$}\\ 0 & other-case\\ \end{cases}$$
So, adding term by term, for every $N>1$, there exists a $k$, $0 \leq k \leq N-1$, such that $2N = p_l + p_m$, where $p_l \leq p_m$, $p_l$ and $p_m$ are primes.
So, for every $N>1$, exits $p_l$ and $p_m$, $p_l \leq p_m$, primes, such that $2N=p_l+p_m$.