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Blazing goIITian

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27 Nov 2008 09:31:17 IST

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Wilson theorem taken ahead...

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Prove that $(p-1)!^{p^{n-1}}\equiv -1\bmod p^n$

Wilson theorem modified....

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Hari Shankar

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Joined: 28 Feb 2007

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27 Nov 2008 13:57:56 IST

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This is easily proved by induction.

In fact it is true for any number a satisfying $a \equiv -1 \bmod {p}$ that $a^{p^{n-1}} \equiv -1 \bmod {p^n}$

To get an idea of how the proof will run lets try proving that if $a \equiv -1 \bmod {p}$ then $a^p \equiv -1 \bmod {p^2}$

We have $a^p +1 =(a+1) (a^{p-1} - a^{p-2}+...-a+1)$

a+1 is divisible by p. And we have to prove that $a^{p-1} - a^{p-2}+...-a+1$ is divisible by p

Note that $a^{p-1} \equiv (-1)^{p-1} = 1 \bmod{p}$ . Next $-a^{p-2} \equiv - (-1)^{p-2} = 1 \bmod{p}$

In this way

$a^{p-1}-a^{p-2}+...-a+1 \equiv \underbrace{1+1+...+1}_{p \ \text{times}} = p \equiv 0 \bmod{p}$ .............................................I

Thus $a^p+1 \equiv 0 \bmod {p^2} \Rightarrow a^p \equiv -1 \bmod{p^2}$

Now the path is clear for using induction.

Let us assume that the hypothesis is true for some number k+1 i.e. $a^{p^k} \equiv -1 \bmod{p^{k+1}}$

and we have to prove that $a^{p^{k+1}} \equiv -1 \bmod{p^{k+2}}$

Now $a^{p^{k+1}} +1 = \left(a^{p^k} \right)^p + 1$ . So let $a^{p^k} = b$

Remember that $b \equiv -1 \bmod {p^{k+1}}$ and also that it automatically follows that $b \equiv -1 \bmod {p}$

So now, $b^p+1 = (b+1) (b^{p-1} - b^{p-2}+...-b+1)$

Already $b+1 \equiv 0 \bmod{p^{k+1}}$

So it remains to prove that $(b^{p-1} - b^{p-2}+...-b+1) \equiv 0 \bmod{p}$ . But this has already been done in I above. And thus $b^p+1 = a^{p^{k+1}}+1\equiv 0 \bmod{p^{k+2}}$

Hence for all $n \in \mathbb{N}$ it is true that $a^{p^{n-1}} \equiv -1 \bmod{p^n}$ whenever $a \equiv -1 \bmod{p}$

In particular, Wilson's Theorem tells us that $(p-1)! \equiv -1 \bmod{p}$ . Hence we have

$(p-1)!^{p^{n-1}} \equiv -1 \bmod{p^n}$

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