Question:
Prove that:among $ 1!, 2!,...,p!$ there are at least $ \sqrt{p}$ different residues in modulo $ p$ where $ p$ is a prime.
This problem it seem hard,Now I can only make $\dfrac{2\sqrt{2p}}{3}$ Would you like share your solution?Thank you
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math110
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Is this something you know to be true, or only conjecture is true? – Thomas Andrews Mar 07 '17 at 03:16
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This problem it's my teacher give me – math110 Mar 07 '17 at 03:19
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how do you get $\frac{2 \sqrt{2p}}{3} \approx 0.94 \sqrt{p}$ ...? – reuns Mar 07 '17 at 03:46
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1As a sidenote: if $N(p)$ is the actual number of different residues then numerical evidence suggest that $\lim_{p\to \infty} \frac{N(p)}{p}$ exists and equals $\simeq 0.6321$. Tested for $2 < p < 10^{7}$. – Winther Mar 07 '17 at 04:04
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@Winther $1-1/e$ perhaps? It would mean that $n!\bmod p$ behaves like a "random" function: https://math.stackexchange.com/questions/2312576 – Bart Michels Nov 30 '17 at 09:10
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I'm still curious how you got $\frac{2\sqrt{2 p}}{3}$. While the constant is worse, the method may be interesting. – Bart Michels Aug 05 '18 at 09:30
1 Answers
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Let $S$ denote the set of non-zero residues expressible as a factorial modulo $p$. Then the function $f\colon S\times S\to \{1,\ldots,p-1\}$ given by $f(x,y)=x/y$ is surjective, because for every $i=2,\ldots,p-1$ we have $f(i!,(i-1)!)=i$, and also $f(1!,1!)=1$. Therefore $|S|^2\geq p-1$. Thus, the number of residues expressible as factorials is $|S|+1\geq \sqrt{p}$ since $(|S|+1)^2\geq |S|^2+1\geq p$.
pre-kidney
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I clarified this, certainly the proof shows that $|S|^2\geq p-1$ because you can obviously get $1$ as $f(1!,1!)$ rather than $f(1!,0!)$ as you correctly pointed out is not included in the set. – pre-kidney Mar 07 '17 at 04:02
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