I've heard all kinds of different ways to solve this problem, yet haven't been able to apply them specifically to the number 8 (Worked fine for 6 for example). I'd love to see a well-explained solution, is possible. Thank you.
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Since $\varphi(p^a)=p^{a-1}(p-1)$ we see that no odd prime can divide $n$ with exponent $>1$ and that any odd prime dividing $n$ must be one greater than a power of $2$. Not hard to check all the cases. – lulu Apr 29 '16 at 12:41
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In general one can compute the inverse of this function. – amcalde Apr 29 '16 at 12:42
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It helps the OP more, if you show how the inverse of this function can be computed. It is not totally trivial. The numbers {$15,16,20,24,30$} satisfy $\phi(n)=8$. – Peter Apr 29 '16 at 12:43
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See http://math.stackexchange.com/a/23955/589 for how hard it is in general. – lhf Apr 29 '16 at 12:44
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Hint: If $p^e$ divides $n$, then $\varphi(p^e)=p^{e-1}(p-1)$ divides $\varphi(n)$. Therefore:
$p=2$ and $e-1 \le 3$, or
$p$ is odd and $e=1$ and $p-1$ divides $8$.
This limits the possible candidates for $p^e$. You then need to argue how these candidates can be combined to yield $\varphi(n)=8$.
lhf
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