solution verification of least residue

Asked Jun 15 '22 at 05:20

Active Jun 15 '22 at 16:03

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I want to find the least residue of $95!$ mod $97$. So I wanna solve $$95! \equiv_{97} x$$ where $x \in \mathbb{Z}^+$ is minimum, but Wilsons theorem

$$96! \equiv_{97} -1$$

so multiplying the top equation by 96 I obtain

$$-1 \equiv_{97} 96x$$

which forces $x=1$. Did I do this correctly using Wilsons theorem?

edited Jun 15 '22 at 16:03

asked Jun 15 '22 at 05:20

homosapien

2

Just to be sure, the $19$'s are meant to be $97,$ right? If so then I agree, but if not then it's clearly false because $19 \leq 95 \Rightarrow 19 | 95! \Rightarrow 95! \equiv_{19} 0$ – Stephen Donovan Jun 15 '22 at 05:31
1

Does this answer your question? Showing $p$ divides $(p-2)!-1$ when $p$ is prime (note that $97$ is prime) - found using an Approach0 search. – John Omielan Jun 15 '22 at 05:56
@StephenDonovan yes edit made. Does it work ? – homosapien Jun 15 '22 at 16:02

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