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$ (n-2)! \equiv 1 \mod n$
If n is said to be a prime number. I guess we'll have to use FERMAT’S LITTLE THEOREM, and I just don't know where to start from. Thanks in advance
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Bledi Boss
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2Related : http://math.stackexchange.com/questions/363738/prove-the-converse-of-wilsons-theorem – lab bhattacharjee Nov 19 '13 at 16:58
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$$n=4;,;;(n-2)!=2!=2\neq 1\pmod 4$$ – DonAntonio Nov 19 '13 at 16:59
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but 4 is not a prime number! – Bledi Boss Nov 19 '13 at 16:59
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Exactly @BlediBoss, so your claim is false unless you require $;n;$ to be a prime. Check my answer. – DonAntonio Nov 19 '13 at 17:00
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@DonAntonio that looks promising :) – Bledi Boss Nov 19 '13 at 17:04
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If $\;n=p\;$ is a prime, then by Wilson's theorem
$$\color{red}{-1}=(p-1)!=(p-2)!(p-1)=\color{red}{-(p-2)!\pmod p}\implies 1= (p-2)!\pmod p$$
DonAntonio
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I forgot to ask, how did you make that pass, I mean how do you eliminate $ (p-1) $ – Bledi Boss Nov 19 '13 at 18:54
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1@BlediBoss, it folllows directly from the definition of modular equality! $;a=b\pmod p\iff p\mid (a-b);$ , so in our case $;p-1=-1\pmod p\iff p\mid(-1-(p-1))\iff p\mid p;$...what, do you think this is trivial or what? :) – DonAntonio Nov 19 '13 at 19:50
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