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I've recently starting reading about number theory, and came across this proposition in A Classical Introduction to Modern Number Theory: problem statement The hint is to start writing $x$ and $a$ as $(-1)^y5^z$ and $(-1)^s5^t$ respectively. For the first case where $n$ is odd, I have:
$x^n \equiv (-1)^{yn}5^{zn} \equiv (-1)^{y2k}(-1)^y5^{zn} \equiv (-1)^y5^{zn}$

where the third step occurs from writing $n = 2k + 1$ for some $k$. Thus we are left with the relation:

$(-1)^y5^{nz} \equiv (-1)^s5^t (2^e)$

but I'm not quite sure where to go from there, any advice would be great, thanks!

Kit
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The hint seems to use the structure of $U(2^e)$.

An elementary reason why $x^n \equiv a \bmod 2^e$ has a unique solution when both $n$ and $a$ are odd is that $x \mapsto x^n$ is a surjective map on $U(2^e)$ and so a bijection.

Indeed, $U(2^e)$ has order $2^{e-1}$. Since $n$ is odd, we have $nu+2^{e-1}v=1$ for some $u,v\in \mathbb Z$. Then $x \equiv (x^u)^n \bmod 2^{e}$.

lhf
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