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visits member for 2 years, 2 months
seen Sep 24 '12 at 21:46

Oct
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Aug
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awarded  Popular Question
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awarded  Popular Question
Sep
24
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 $p^{n+1}$?
Sep
24
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.
Sep
24
comment Show $[(p-1)!]^{p^{n-1}} \equiv -1 $ (mod $p^n$) for n $\in \mathbb N$
@ThomasAndrews Oh! That makes sense.
Sep
24
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.
Sep
24
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?
Sep
24
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.
Sep
18
accepted Characterising reals with terminating decimal expansions
Sep
18
asked Characterising reals with terminating decimal expansions
Sep
18
accepted Fourier transform of inverse rectangular pulse
Sep
18
answered Fourier transform of inverse rectangular pulse
Sep
18
accepted If $a^2$ divides $b^2$, then $a$ divides $b$
Sep
18
accepted $p$ a prime, $p \equiv 3 \pmod 4$. Prove that $\frac{p-1}{2}! \equiv \pm 1 \pmod p$
Sep
18
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!
Sep
18
comment $p$ a prime, $p \equiv 3 \pmod 4$. Prove that $\frac{p-1}{2}! \equiv \pm 1 \pmod p$
In the general case, $\frac{p-1}{2}$ isn't odd though? e.g. p = 17