Wilson's theorem related problem

prove $$18! \equiv -1 \pmod{437}$$

I do not want full solution to the above problem but if anybody can tell me how we can approach to it, I will really appreciate that.

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Note $437 = 19\cdot 23$. – Sean Eberhard Aug 26 '12 at 9:26

hints :

• $437=19\cdot 23$ (as proposed by Sean)
• Wilson's theorem :-)
• $19\cdot 20\cdot 21 \cdot 22=(-4)(-3)(-2)(-1)\pmod{23}$
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As noted by Sean, $\,427=19\cdot 23\,$, thus using Wilson' theorem twice: $$(1)\,\,\,18!\cdot 19\cdot 20\cdot 21\cdot 22=22!=-1\pmod {23}\Longrightarrow$$ $$\Longrightarrow 18!=\frac{-1}{(-4)(-3)(-2)(-1)}=-\frac{1}{24}=-1\pmod {23}$$ $$(2)\,\,\,\,18!=-1\pmod {19}$$

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Is this not a full solution? – awllower Aug 26 '12 at 10:43
Not yet...but almost. The gist of it all is number (1), as (2) is directly Wilson's Theorem – DonAntonio Aug 26 '12 at 10:48
Indeed, this cqn be interpreted this way!! – awllower Aug 26 '12 at 10:49