<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] Here's a more elementary argument.  Note that mod $10$ we have
$$a^{4n+1}-a = a(a^{4n}-1) = a(a^{2n}-1)(a^{2n}+1) \equiv a(a^{2n}-1)(a^{2n}-9) =a(a^n-1)(a^n+1)(a^n-3)(a^n+3) \pmod{10}.$$
If $\gcd(a,10)=1$ or $10$, then $a^n \equiv 0, \pm1 \mbox{ or }\pm3 \pmod{10}$, making the above product congruent to $0$ mod $10$.  
If $\gcd(a,10) = 5$, then $a$ is odd and each of the other factors in the product is even, so again the product is congruent to $0$ mod $10$.
If $\gcd(a,10) = 2$, then $a^n \equiv 2, 4, 8, \mbox{ or } 6 \pmod{10}$  Each of these is $\pm 1$ or $\pm 3$ from $5$, so $5$ is a factor of the product and again the product is congruent to $0$ mod $10$.
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