text: # last digit of $n^5$ and $n$ is the same digit [duplicate]
  
  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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  ## marked as duplicate by lhf, M Turgeon, Douglas S. Stones, Steven Stadnicki, Pedro Tamaroff♦Sep 25 '12 at 1:56
  
  This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
  
  This is answered in the following answer to the linked question: math.stackexchange.com/a/184623/19379 – M Turgeon Sep 25 '12 at 1:24
  If it's the same question, why ask again? – lhf Sep 25 '12 at 1:28
  I think a relation holds: $n^{4k + 1} \equiv n \pmod{10}$ for any positive integer $k$. – Amad27 Jun 30 '15 at 10:14
  
  ## 2 Answers
  
  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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  Don't $a$ and $n$ have to be coprime ? – Belgi Sep 25 '12 at 1:26
  Ahh yes that was very clear. Thank you. – MinaHany Sep 25 '12 at 1:31
  
  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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  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 1:37
  That was not true in general; i removed it. – Aang Sep 25 '12 at 1:44
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