<s>[INST] Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit 
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit

I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
 [/INST] The most elementary method is simply to observe patterns of the last digits ( there are only 10 possible digits to check so it is quite fast)  by multiplying last digit by $a,a^2,a^3,a^4 $, for greater powers - the pattern is repeating ..
$0 \rightarrow 0,0,0,0,.. $
$1 \rightarrow 1,1,1,1,.. $
$2 \rightarrow 4,8,6,2,.. $
$3 \rightarrow 9,7,1,3,.. $
$4 \rightarrow 6,4,6,4,.. $
$5 \rightarrow 5,5,5,5,.. $
$6 \rightarrow 6,6,6,6,.. $
$7 \rightarrow 9,3,1,7,.. $
$8 \rightarrow 4,2,6,8,.. $
$9 \rightarrow 1,9,1,9,.. $   
From this you can devise even the more general proposition - if $a$ and $b$ have the same last digit then $ab^{4m}$ or $ba^{4m}$ have the same last digit like $a$ and $b$.
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