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Someone could help me find some error in the reasoning:

We know, that the canonical decomposition of $n!$ is:

$n!=\prod_{p_{i}\leq n}p_{i}^{\alpha_{i}(n)}$, where:

$\alpha_{i}(n)=\sum_{t=1}^{r}[\frac{n}{p_{i}^{t}}]$.

It's possible use the canonical decomposition of $n!$ to found the canonical decomposition of $n$, let's see:

$n=\frac{n!}{(n-1)!}=\frac{\prod_{p_{i}\leq n}p_{i}^{\alpha_{i}(n)}}{\prod_{p_{i}\leq n-1}p_{i}^{\alpha_{i}(n-1)}}=\prod_{p_{i}\leq n}p_{i}^{\alpha_{i}(n)-\alpha_{i}(n-1)}=\prod_{p_{i}\leq n}p_{i}^{\beta_{i}(n)}$, 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:

$\forall N > 1$, $\exists k\in \{0,1,2,...,N-1\}$ such that:

$N^{2}-k^{2}=\prod_{p_{i}\leq N^{2}-k^{2}}p_{i}^{\gamma_{i}(N,k)}$, and:

$\gamma_{i}=\begin{cases}{1}&\text{if}& i=l\\1 & \text{if}& i=m\\0 & \text{if}& other-case\end{cases}$

This is where I need help to find some error in the reasoning:

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:

$(N-k)(N+k)=\prod_{p_{i}\leq N+k}p_{i}^{\beta_{i}(N-k)+\beta_{i}(N+k)}$

$N^{2}-k^{2}=\prod_{p_{i}\leq N+k}p_{i}^{\gamma_{i}(N,k)}$, where:

$\gamma_{i}(N,k)=\beta_{i}(N-k)+\beta_{i}(N+k)$, then:

$\gamma_{i}(N,k)=\alpha_{i}(N-k)-\alpha_{i}(N-k-1)+\alpha_{i}(N+k)-\alpha_{i}(N+k-1)$

The only way (due to the uniqueness of the decomposition), to satisfy:

$\gamma_{i}=\begin{cases}{1}&\text{if}& i=l\\1 & \text{if}& i=m\\0 & \text{if}& other-case\end{cases}$

is to suppose that $N-k=p_{l}$ and $N+k=p_{m}$ for some $p_{l}< 2N$, $p_{m}< 2N$ primes, lets see:

$\gamma_{i}(N,k)=\alpha_{i}(N-k)-\alpha_{i}(N-k-1)+\alpha_{i}(N+k)-\alpha_{i}(N+k-1)$

$\gamma_{i}(N,k)=\alpha_{i}(p_{l})-\alpha_{i}(p_{l}-1)+\alpha_{i}(p_{m})-\alpha_{i}(p_{m}-1)$

$\gamma_{i}(N,k)=\beta_{i}(p_{l})+\beta_{i}(p_{m})$

$\gamma_{i}(N,k)=\delta_{i,l}+\delta_{i,m}=\begin{cases}{1}&\text{if}& i=l\\1 & \text{if}& i=m\\0 & \text{if}& other-case\end{cases}$

(Q.E.D.)

Then:

$\forall N > 1$, $\exists k\in \{0,1,2,...,N-1\}$ such that:

$N^{2}-k^{2}=\prod_{p_{i}\leq N^{2}-k^{2}}p_{i}^{\gamma_{i}(N,k)}$, and:

$\gamma_{i}=\begin{cases}{1}&\text{if}& i=l\\1 & \text{if}& i=m\\0 & \text{if}& other case\end{cases}$

That is to say:

$N^{2}-k^{2}=\prod_{p_{i}\leq N^{2}-k^{2}}p_{i}^{\gamma_{i}(N,k)}$

$N^{2}-k^{2}=p_{l}^{\gamma_{l}}p_{m}^{\gamma_{m}}$

$N^{2}-k^{2}=p_{l}p_{m}$

$(N-k)(N+k)=p_{l}p_{m}$, then:

$N-k=p_{l}$ and $N+k=p_{m}$

Finanlly, adding the equations:

$2N=p_{l}+p_{m}$

That is to say: "Any even number greater than two can be expressed as the sum of two prime numbers."

Thanks for your help.

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    $\begingroup$ You do realise that $N^2-k^2$ is either odd or divisible by $4$? This means that $\gamma_1$ can never be 1, contrary to your "hypothesis". $\endgroup$ – N. S. Dec 5 '18 at 17:04
  • $\begingroup$ $10^{2}-3^{2}=7*13$, $10^{2}-7^{2}=3*17$, $11^{2}-0^{2}=11*11$, $11^{2}-6^{2}=5*17$, $11^{2}-8^{2}=3*19$ $\endgroup$ – Mauricio Areiza Dec 5 '18 at 17:22
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    $\begingroup$ What are $m,l$ in the definition of $\gamma_i$? $\endgroup$ – Alex R. Dec 5 '18 at 19:11
  • $\begingroup$ @AlexR. If $l$, let's say, is equal to $2$ ($l=2$), then $\gamma_{2} $ is the exponet asosiated to $p_{2}=3$ (the second prime number). $\endgroup$ – Mauricio Areiza Dec 5 '18 at 19:24
  • $\begingroup$ What confuses me is your reasoning for $\gamma_i$ in terms of $\beta_i$. It's certainly true that $\beta_i(p_j)=\delta_{ i,j}$ (but certainly not true for $\beta_i(n)$ for $n$ not prime), but you're concluding $\gamma_i=1$ by first implying that you've found a prime $N-k=p_l$ to force $\gamma_l=1$ and therefore there's a prime $p_m=N+k$. What I'm saying is, your proof that $N-k=p_l$ is cyclical because you assume that such primes $p_l,p_m$ exists in the first place. It's true that if you can show $N^2-k^2 = p_lp_m$ for some primes $p_l,p_m$ that would imply the Goldbach conjecture. $\endgroup$ – Alex R. Dec 5 '18 at 19:51

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