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I have got a question which is as follows:

If $p$ is a prime number, show that $2(p-3)!+1$ is a multiple of $p$.

I know that this question can be solved using Wilson theorem which is the only thing I apply in such situations, but I don't know how can I use it here.

Please help.

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You have got it right, the use of Wilson theorem suffices here.

By Wilson theorem, $p\mid (p-1)!+1$ (Where $p$ is prime);

i.e.,
$$ \begin{align}p&\mid 1+(p-1)(p-2)(p-3)!\\ \implies p&\mid(p^2-3p+2)(p-3)!+1\\ \implies p&\mid 1+(p^2-3p)(p-3)!+2(p-3)!\\ \implies p&\mid 1+p(p-3)(p-3)!+2(p-3)!\\ \implies p&\mid 1+2(p-3)! \end{align} $$ As $p\mid p(p-3)(p-3)!$

This implies $1+2(p-3)!$ is a multiple of $p$.

I think that is what we needed to prove.

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  • $\begingroup$ What about $p=2$? :) $\endgroup$ – EzWin Dec 22 '16 at 16:09
  • $\begingroup$ Look at jyraki's comment, he has said very much in few words $\endgroup$ – I am Back Dec 22 '16 at 16:20
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Hint: Multiply your number by $(p-2)(p-1)$. Then use Wilson theorem.

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    $\begingroup$ And observe that you need the assumption $p>2$ at selected steps :-) $\endgroup$ – Jyrki Lahtonen Dec 22 '16 at 14:24
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$(p-1)! \equiv -1 \mod p$ (Wilson's theorem). So, $(p-3)!(p-2)(p-1) \equiv (p-3)!(-2)(-1) \equiv 2*(p-3)! \equiv -1 \mod p$. So, $2*(p-3)!+1 \equiv 0 \mod p$

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