So, I was thinking that $a^{\phi(m)}\equiv 1 \bmod{m}$, thus when multiplying $a^t$ on both sides, we get that $a^{\phi(m)+t} \equiv a^t \bmod{m}$. What is throwing me off is the all integers a part.
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$$a^{\phi(p^t)+t}-a^t=a^t(a^{\phi(p^t)}-1)$$
Now as $p$ is prime; either $p$ divides $a$ or $p\not|a\iff (p,a)=1$
If $p$ divides $a, p^t|a^t$
Else $(p,a)=1$ which can safely use Euler's Totient Theorem
lab bhattacharjee
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1Correct me if I am wrong, but is the $a^{\phi(p^t) + t} - a^{\phi(t)}$ supposed to be $a^{\phi(p^t) + t} - a^t$? – terminix00 Apr 15 '14 at 06:13
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1@user2977382, Sorry for the typo – lab bhattacharjee Apr 15 '14 at 06:14
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So If I am understanding this correctly, if $p^t \mid a^t$ then we get $a^{\phi(t)}*a^t - a^t \equiv 0 \bmod{p^t}$ and subsequently $0 \equiv 0 \bmod{p^t}$ which is true? – terminix00 Apr 15 '14 at 06:34
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1@user2977382, Agreed – lab bhattacharjee Apr 15 '14 at 06:35
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1Thanks a lot. Do you live on this site? I see you everywhere. – terminix00 Apr 15 '14 at 06:36
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@user2977382, My pleasure. Could I make the idea clear – lab bhattacharjee Apr 15 '14 at 12:06