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Question: Let $p$ be prime, and suppose that $r$ is a positive integer less than $p$ such that $(−1)^rr! ≡ −1($mod $ p)$. Show that $(p − r + 1)! ≡ −1 ($mod $ p)$.

I'm not really sure what to do. My first strategy was to take $(−1)^rr!$, and expand it out. Doing so, we can get: $(-r)(-r+1)(-r+2)...(-2)(-1)$ My next assumption was to just add $p$ to all of these terms, but that doesn't get us what we want. Ultimately, it seems like we want to show $(p-r+1)(p-r)(p-r-1)(p-r-2)(2)(1)\equiv(−1)^rr!$ . There are more terms though potentially in $(p-r+1)!$ than $(−1)^rr!$, so I'm not really sure what to do. Perhaps connecting Wilson's theorem with the given? Hints would be greatly appreciated. Thanks!

Edit: Here's a screenshot directly from the book I am using.

The book is Elementary Number Theory, 6th edition by Kenneth H. Rosen.

enter image description here

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    $\begingroup$ Let $r=3,p=5$ and note that $(-1)^3\times3!=-6\equiv -1\pmod 5$ but $(5-3+1)!\equiv 3!\equiv 6\equiv 1\pmod 5$...so something seems off. $\endgroup$ – lulu Nov 25 '18 at 21:22
  • $\begingroup$ I'm not sure! I posted a screenshot of the exact question. $\endgroup$ – Stawbewwy Nov 25 '18 at 21:28
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    $\begingroup$ Then there is a mistake in the exercise. It happens. $\endgroup$ – Jean-Claude Arbaut Nov 25 '18 at 21:28
  • $\begingroup$ Well, I don't think I botched my counterexample (though of course I might have). $\endgroup$ – lulu Nov 25 '18 at 21:29
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    $\begingroup$ As another counterexample: $r=4,p=5$. We check that $(-1)^4\times 4!=24\equiv -1 \pmod 5$ but $(5-4+1)!=2!=2\pmod 5$. $\endgroup$ – lulu Nov 25 '18 at 21:30
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Are you sure that what you are trying to prove is even true? Below you can find a counterexample to your problem.

Let $p$ be a prime and let $r=1$. Then it is always the case that $(-1)^rr!\equiv-1\bmod p$, but $(p-r+1)!\equiv p!\equiv 0 \bmod p$.

Just in case anyone wants a second counterexample, since $r=1$ feels kind of silly...

Let $p$ be a prime greater than $3$, and let $r=p-1$. Then it is always the case that $(-1)^rr!\equiv(p-1)!\equiv-1\bmod p$ by Wilson's Theorem, but $(p-r+1)!\equiv 2!\equiv 2 \bmod p$.

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