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Possible Duplicate:
The last digit of $n^5-n$
Why is the last digit of $n^5$ equal to the last digit of $n$?

Basically, this is the same question as Why is the last digit of $n^5$ equal to the last digit of $n$?

What I want to prove is

$n^5 ≡ n$ mod 10

Since I'm studying Euler's Phi Function, I know that the proof of this is related to it. So I'm looking to prove this using the Phi function. A comment on the original question suggests $φ(10)=4$ but I don't see how I can use this. Anyone can point me in the right direction? Thanks

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    I think a relation holds: $n^{4k + 1} \equiv n \pmod{10}$ for any positive integer $k$.2015-06-30

2 Answers 2

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Euler's theorem states that if $a$ and $n$ are relatively prime, then

$a^{\varphi(n)} ≡ 1 \mbox{ mod } n$

Multiply by $a$ on both sides,

$a^{\varphi(n) + 1} ≡ a \mbox{ mod } n$

Now set $n = 10$. Does this help?

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    Ahh yes that was very clear. Thank you.2012-09-25
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When $n\equiv 0,1\pmod 2\implies n^5\equiv 0,1\pmod 2\implies n^5\equiv n\pmod {2}$

Also, $n^5\equiv n\pmod {5}$ (By Fermat's little theorem)

By Chinese remainder theorem,

These both $\implies n^5-n\equiv 0\pmod {10}\implies n^5\equiv n\pmod {10}$

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    That was not true in general; i removed it.2012-09-25