Twin Primes- Clement conjecture proof For a work I'm doing, I need to provide the proof of the Clement theorem
$$
\text{($n$ and $n+2$ are twin primes)} \quad\Longleftrightarrow\quad 4[(n-1)!+1]+n \equiv 0 \pmod{n(n+2)}.
$$
So I decided to look for the original article, which i provide here https://www.jstor.org/stable/2305816?origin=crossref&seq=1#metadata_info_tab_contents. I understood the necessity correctly, but the sufficiency I had some trouble with. On the article it says it's obvious because we can reduce it by divisions to a modified version of the Wilson theorem. So I did
$$\begin{aligned}
4[(n-1)!+1]+n \equiv 0 \pmod{n(n+2)}
&\quad\Rightarrow\quad 4[(n-1)!+1]+n \equiv 0 \pmod{n}\\
&\quad\Rightarrow \quad 4[(n-1)!+1] \equiv 0 \pmod{n} \quad\text{(1),}
\end{aligned}
$$
but this doesn't mean that
$$[(n-1)!+1] \equiv 0 \pmod{n},$$ also for example $n=4$ verifies (1), but it's not a prime. Anyone knows what was the article referring to?
P.S. sorry if i made some grammar mistakes, I'm not that good at English.
 A: Let, $T=4[(n-1)!+1]+n$. If $n(n+2)|T$ then $n|4[(n-1)!+1]$ which according to Wilson's theorem requires $n$ to be a prime.
Because, if $n=2a$, $a$ odd, then $a|(2a-1)!+1$. But, $a|(2a-1)!$. Similar thing happens for $n=4a$. Hence,  $n|(n-1)!+1$ which implies by Wilson's theorem that $n$ is a prime.
Now, $\text{gcd}(n,n+2)=1$ as $n>2$ and $n$  is a prime and $n+2$ is odd.
Let, $n+2=p$. $(n+2)|4[(n-1)!+1]+n \Rightarrow (n+2)|2(n-1)!+1 $ or $p|2(p-3)!+1$
But, $(p-1)!+1=p^2(p-3)!-3p(p-3)!+[2(p-3)!+1]$
If $p|2(p-3)!+1$ then $p|(p-1)!+1$, which according to Wilson's theorem implies that $p=n+2$ must be a prime number. Hence, $n, n+2$ must be twin primes.
The converse is much straightforward to prove.
A: If $n$ is odd, you can do the desired division. If $n$ were even, $n=2k$ and $n+2=2(k+1)$, and since $4(n-1)!$ is divisible by $4k(k+1)=n(n+2)$ you can quickly derive a contradiction.
A: It follows immediately (mechanically!) by $\rm\color{#90f}W$ = Wilson's theorem and (Easy) CRT as below,
using $\!\bmod n\!+\!2\!:\ \color{#0a0}{(n\!+\!1)!} = \smash{\underbrace{(n\!+\!1)n}_{\large \ \ (-1)(-2)}\!\!(n\!-\!1)!\equiv \color{#0a0}{2(n\!-\!1)!}}\,\ $ [Wilson reflection]
$\ \begin{align}
n\!+\!2\ {\rm prime} &\smash{\overset{\small \rm\color{#90f} W\!}\iff} \overbrace{\color{#0a0}{(n\!+\!1)!}}^{\large\color{#0a0}{2(n-1)!}}\equiv -1\!\!\!\pmod{\!n\!+\!2}\smash{\overset{\color{#0a0}{\times\ 2}}\iff} 4(n\!-\!1)!\equiv -2\!\!\!\pmod{\!n\!+\!2}\\[.5em]
\&\ \ \, n\ {\rm prime} &\smash{\overset{\small \rm\color{#90f} W\!}\iff} (n\!-\!1)!\equiv -1\!\!\!\pmod{\!n}\ \ \ \smash{\overset{\times\ 4}\iff}\ \ \ 4(n\!-\!1)!\equiv -4\!\!\!\pmod{\!n}\\
\end{align}$
$$\,\ \overset{\small \rm CRT}\iff 4(n\!-\!1)!\equiv -4 + n\underbrace{\left[\dfrac{2}{\color{#c00}n}\bmod n\!+\!2\right]}_{\large \color{#c00}n\ \equiv\ -2}\equiv\,\bbox[5px,border:1px solid #c00]{-4\!-\!n}\ \pmod{\!n(n\!+\!2)}$$
Clearly the same method generalizes to any prime $k$-tuple.
