Solve it

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Rabeeb
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Solve it

Unread post by Rabeeb » Thu Feb 28, 2013 1:43 pm

If $(p-1)! \equiv -1 \text{ or } 0 \pmod{p}$ is false,
Find the values of $p$

Swargo
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Re: Solve it

Unread post by Swargo » Thu Sep 05, 2013 10:05 pm

প্রশ্নটা ঠিকভাবে লিখলে ভাল হয়। বোঝা যাচ্ছে না।

Nayeemul Islam Swad
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Re: Solve it

Unread post by Nayeemul Islam Swad » Mon Feb 03, 2014 3:54 pm

How can it be congruent to $0$!!! How can there exist such primes either!!! :!:
Why so SERIOUS?!??!

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*Mahi*
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Re: Solve it

Unread post by *Mahi* » Mon Feb 03, 2014 8:49 pm

Read the question carefully. There is no mention of $p$ being prime.
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

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asif e elahi
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Re: Solve it

Unread post by asif e elahi » Tue Feb 04, 2014 10:58 am

Rabeeb wrote:If $(p-1)! \equiv -1 \text{ or } 0 \pmod{p}$ is false,
Find the values of $p$
$p=4$ is the only solution. If $p$ is a prime,then by Wilson's theorem $(p-1)!\equiv -1(mod p)$. If $p$ is composite,we can write $p=ab$ where $a,b>1$.If $a=b$ and $a>2$,then $a<2a<a^{2}-1=p-1$.So $a\times 2a=2p\mid (p-1)!$.This implies $(p-1)!\equiv 0(mod p)$.If $a>b$,then $a>b>p-1$. So $p=ab$ divides $(p-1)!$ or $(p-1)!\equiv 0(mod p)$. So $p=4$ is the only solution.

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