Practice Problems For Choosing mod

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photon
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Re: Practice Problems For Choosing mod

Unread post by photon » Sat Feb 04, 2012 3:18 pm

Tahmid Hasan wrote: 5.let $p^k+1=t^2$
let $p=2$,then $2^k=(t+1)(t-1)$
...................
if $p$ is odd then $p^k=(t+1)(t-1)$,$(t+1,t-1)=1$ since both are odd
i think it is wrong.taking the lesser one as 1,t-1=1 :arrow: t=2
then,$(p,k)=(3,1)$
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FahimFerdous
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Re: Practice Problems For Choosing mod

Unread post by FahimFerdous » Sat Feb 04, 2012 5:57 pm

Photon, what Tahmid wrote is correct. But I'm nt understanding where saw the bug and what you wrote. Why (p,k)=(3,1)?
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Re: Practice Problems For Choosing mod

Unread post by sourav das » Sun Feb 05, 2012 11:10 am

Actually Tahmid stop working showing $(t+1,t-1)=1$. But the case $t-1=1,t+1=p^k$ was the missing case.So indeed $(p,k)=(3,1)$ is solution.
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Re: Practice Problems For Choosing mod

Unread post by sm.joty » Sun Feb 05, 2012 4:55 pm

What is LTE ????? please give a link or explain it.
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nafistiham
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Re: Practice Problems For Choosing mod

Unread post by nafistiham » Sun Feb 05, 2012 5:46 pm

it is Lifting the exponent lemma.
[a very useful lemma, which i have not use fruitfully yet :oops: :cry: ]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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