Discussion on Bangladesh Mathematical Olympiad (BdMO) National
Sudip Deb new
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Unread post by Sudip Deb new » Sun Jan 30, 2011 10:04 am

Prove that (n^5 - n) is divided by 5 .

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Re: Prove

Unread post by HandaramTheGreat » Sun Jan 30, 2011 10:18 am

last digit of a square can be only 1, 4, 9, 6, 5 or 0...
if it's 1, 4, 9, or 6, then $(n^2+1)$ or $(n^2-1)$ is obviously divisible by 5...
if it's 5 or 0, then $n$ should be divisible by $5$... :D
if you put 2 dollar sign in both side of (n^5-n), it would be like this $(n^5-n)$... ;)

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Re: Prove

Unread post by Moon » Sun Jan 30, 2011 7:47 pm

Also you can always use Fermat's Little Theorem, which states that for all $n\in\mathbb{N}$ and any prime $p$ \[ n^p \equiv n \pmod{p} \]
When $gcd(n,p)=1$, you can divide the equivalence by $n$ and get $n^{p-1} \equiv 1 \pmod{p}$.
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