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Let $p \geq 3$ be a prime number.

How to prove that $(p-3)!+2^{p-2} \equiv 0 \ (\mathrm{mod} \ p)$?

My idea was to use Wilson's theorem:

Since $p$ is a prime it follows that $(p-1)! \equiv -1 \ (\mathrm{mod} \ p)$.

$\Rightarrow 1 \cdot \cdot \ \cdot (p-2)(p-3)+2^{p-2} \equiv -1 \ (\mathrm{mod} \ p)$

Then $1 \cdot \cdot \ \cdot (p-2)(p-3)+2^{p-2} +1 \equiv 0 \ (\mathrm{mod} \ p)$

Now I'm not sure what to do next. Is this way right or how can it be shown correctly?

Gerturter
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2 Answers2

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By Wilson, $$-1\equiv(p-1)!=(p-1)(p-2)(p-3)!\equiv 2\,(p-3)!\pmod p$$ and by Fermat $$1\equiv 2^{p-1}=2\times 2^{p-2}\pmod p.$$ So $$0\equiv2((p-3)!+2^{p-2})\pmod p.$$

Especially Lime
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Angina Seng
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Just muck.

Bearing in mind $p-2,p-1 \equiv 2, 1 \pmod p$ and the $\pm2^{p-1} \equiv \pm1\pmod p$ and that $\gcd(2,p)=1$ so $2^{-1}\pmod p$ exists and is meaningful.....

$(p-1)! \equiv (p-3)!(p-2)(p-1)\equiv (p-3)!\cdot 2\pmod p$.

And noting Wilsons Th. $(p-1)! \equiv -1\pmod p$ then

$(p-3)!\cdot 2 \equiv -(2^{p-1})\pmod p$

$(p-3)!\equiv -(2^{p-1})\cdot 2^{-1} \equiv -2^{p-2} \pmod p$ and that's it

$(p-3)! +2^{p-2} \equiv 0 \pmod p$ we are done.

What does this mean? Well, if it means anything significant I missed it as all I did was muck about with well, known identities.

fleablood
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