confused
Reputation
226
Next privilege 250 Rep.
 Feb4 awarded Popular Question Oct15 awarded Famous Question Aug11 awarded Popular Question Jul2 awarded Curious Feb9 awarded Yearling Feb6 awarded Notable Question Sep29 awarded Popular Question Sep24 comment Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$ Thanks. @martini I can follow it better now. In the term, $k^2p^{2n}\sum_{l=2}^p \binom pl (kp^n)^{l-2}(-1)^{p-l}$, wouldn't some of the terms of this sum be non-integer fractions, since $\binom pl$ can be a non-integer fraction? Won't this create a problem when working modulo $p^{n+1}$? Sep24 comment Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$ Showing that $(x^{p-1}-x^{p-2}+x^{p-3}...)$ is divisible by p, doesn't look any easier of a task. Sep24 comment Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$ @ThomasAndrews Oh! That makes sense. Sep24 comment Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$ Am I able to do it without the binomial though? The notes that I saw this problem in don't touch on binomial expansion. Sep24 comment Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$ Thanks @Thomas. how do you derive this result? Sep24 comment Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$ I don't really understand the use of combinations like this. Sep18 accepted Characterising reals with terminating decimal expansions Sep18 asked Characterising reals with terminating decimal expansions Sep18 accepted Fourier transform of inverse rectangular pulse Sep18 answered Fourier transform of inverse rectangular pulse Sep18 accepted If $a^2$ divides $b^2$, then $a$ divides $b$ Sep18 accepted $p$ a prime, $p \equiv 3 \pmod 4$. Prove that $\frac{p-1}{2}! \equiv \pm 1 \pmod p$ Sep18 comment $p$ a prime, $p \equiv 3 \pmod 4$. Prove that $\frac{p-1}{2}! \equiv \pm 1 \pmod p$ Oh! but it is, when p $\equiv$ 3(mod 4). I see!