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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 $437 = 19\cdot 23$.2012-08-26

3 Answers 3

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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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    Indeed, this cqn be interpreted this way!!2012-08-26
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Wilson's Theorem: (p-1)! = -1 (mod p), p: is a prime number

Note: When I write the "equal" (=) sign, I mean the congruence sign.....

437 is not a prime number: 437 = 19*23

For: 19 By Wilson's Theorem => 18! = -1 (mod 19)

For 23: By W.T => 22! = -1 (mod 23) => 22*21*20*19*18! = -1 (mod 23)

For 22 = -1 (mod 23) For 21 = -2 (mod 23) For 20 = -3 (mod 23) For 19 = -4 (mod 23)

We multiply our reminders => (-1)(-2)(-3)*(-4) = 24

Back to our equation: 22*21*20*19*18! = -1 (mod 23) transformed to => (-1)(-2)(-3)*(-4) = 24 * 18! = -1 (mod 23)

For 24 = 1 (mod 23) (replace 24 by 1)

1 * 18! = -1 (mod 23)

So, 18! = -1 (mod 23)

Hope it helps you understand it more.