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.
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5$\begingroup$ Welcome to MathOverflow. Grammar and spelling errors from non-native English speakers are treated leniently on MO, and what you've written is fine. However, your mathematical formatting made it very hard to read your question. I've edited it to display a lot of your equations, which will make it easier for people to read. $\endgroup$– Joe SilvermanApr 11, 2021 at 22:28
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$\begingroup$ Just for curiosity, there is a generalization of Clement's theorem that appeared in a 2010 iberoamerican university competition (see ciim.uan.edu.co/ciim-2010-pruebas problem 5). Namely, let $n,d > 1$ be integers with $\gcd(n,d!)=1$. Then $n$ and $n+d$ are both primes if and only if $d!d((n-1)!+1)+n(d!-1) \equiv 0 \pmod {n(n+d)}$. $\endgroup$– SávioApr 12, 2021 at 14:10
3 Answers
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.
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.
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.
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$\begingroup$ $n$ is odd by $n,n\!+\!2$ twins, so $2$ is a unit mod $n$ & $n\!+\!2$, so scaling by $\color{#0a0}2$ & $4$ yields equivalent congruences. $\endgroup$ Jun 19, 2022 at 14:48