last digit of $n^5$ and $n$ is the same digit

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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

This is answered in the following answer to the linked question: http://math.stackexchange.com/a/184623/19379 — M Turgeon, Sep 25 '12 at 01:24
I think a relation holds: $n^{4k + 1} \equiv n \pmod{10}$ for any positive integer $k$. — Amad27, Jun 30 '15 at 10:14

score 4 · Accepted Answer · answered Sep 25 '12 at 01:21

4

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?

answered Sep 25 '12 at 01:21

Elchanan Solomon

30,807

3

Don't $a$ and $n$ have to be coprime ? – Belgi Sep 25 '12 at 01:26
Ahh yes that was very clear. Thank you. – DewinDell Sep 25 '12 at 01:31

Aang · Answer 2 · 2012-09-25T01:43:51.097

4

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

edited Sep 25 '12 at 01:43

answered Sep 25 '12 at 01:27

Aang

14,672

Can you please add detaild on your third paragraph ? Why is the condition about $a$ and $n$ being coprime can be remived ? – Belgi Sep 25 '12 at 01:37
That was not true in general; i removed it. – Aang Sep 25 '12 at 01:44

last digit of $n^5$ and $n$ is the same digit

2 Answers2