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I am currently working on this problem: Prove that if p is prime, then among $1!,2!,\dots,(p-1)!$ one can find at least $\frac{p-1}{2}$ different remainders modulo $p$. I don't really have any clue how to prove it. Can anyone help me?

atdaotlohbh
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    What have you tried? Please use MathJax. Here is a tutorial. Also search this site - start with this post. – Dietrich Burde Feb 18 '24 at 11:04
  • I tried to use primitive roots, but it didn't really help me. Another my idea was that if $a!=(a+k)!$, then $(a+1) \dots (a+k)=1$, which is a polynomial of degree k, so by Lagrange it has not more than k roots. Knowing also that $a!=(a+k)!$ makes sense only for $a=1,\dots ,p-1-k$, we get that $a!=b!$ can not occur for more that $\frac{p^2-1}{4}$ pairs $a,b$. But it gives us a very weak bound on the number of remainders - just over 2 for big primes $p$. – atdaotlohbh Feb 18 '24 at 11:33
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    What is the source of this problem? It looks like an open conjecture to me. You can only prove weaker bounds, it seems. If you don't have "any clue", you should provide more context for us. – Dietrich Burde Feb 18 '24 at 11:33
  • I was just playing with factorials mod primes, and asked myself how many remainders can I get. I wrote some code in python to get the answer, and it seems like it is around $0.63p$. I decided to try to prove $\frac{p}{2}$ bound, worked on it a day and didn't really get anything. And now I posted it here. Sorry, if I somehow break the rules or policy of the site, it is not intentional – atdaotlohbh Feb 18 '24 at 11:42
  • No problem. It is just that some people don't really explain how and where they came to the problem, but one get's much better answers if one gives this context. Some people will not really believe that you were "just playing with factorials mod primes". It seems to be a well known open problem. Perhaps it is part of a "challenge" somewhere. – Dietrich Burde Feb 18 '24 at 11:49

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The conjecture is that the number $N(p)$ of different remainders modulo $p$ among these factorials $1!,\ldots (p-1)!$, satisfies $$ \lim_{p\to \infty} \frac{N(p)}{p}=1-\frac{1}{e}\sim 0.6321205588285576784 $$ So $N(p)\ge \frac{p-1}{2}$ for $p$ large enough. For $p\le 10^7$ this bound is true by a computer verification - see the duplicate here:

among $ 1!,2!,...,p!$ there are at least $ \sqrt{p}$ different residues in modulo $ p$

Dietrich Burde
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