<s>[INST] Find all solutions to $x^{10} = 1 \pmod {377}$ Find all solutions to $x^{10} \equiv 1 \pmod {377}$
I noticed that $x^{10} \equiv 1 \pmod {377}$ can be written as: 
$(x^5+1)(x^5-1)\equiv 0 \pmod {377}$
 also $377 = 13 \times 29$
Any help would be greatly appreciated 
 [/INST] Hint: by the Chinese remainder theorem, we can find all solutions to this equation by first finding all solutions to
$$
x^{10} \equiv 1 \pmod{13}\\
x^{10} \equiv 1 \pmod{29}
$$
To find the solutions mod $13$, it is slightly useful to note that $x^{10} = x^{12 - 2} = x^{-2}$.

Full solution: The first equation has solution
$$
x^{10} \equiv 1 \pmod{13} \iff \\
x^{-2} \equiv 1 \pmod{13} \iff \\
x^{2} \equiv 1 \pmod{13} \iff \\
x \equiv \pm 1 \pmod{13}
$$
so, we have two solutions to the first equation.  For the second, the same trick looks a little weirder:
$$
x^{10} \equiv 1 \pmod{29} \implies\\
(x^{10})^3 \equiv 1 \pmod{29} \implies\\
x^{30 - 28} \equiv 1 \pmod{29} \implies\\
x^2 \equiv 1 \pmod{29}
$$
so, in order for $x$ to be a solution to the original equation, we must also have $x^2 \equiv 1 \pmod{29}$.  So, our full solution is
$$
x \equiv \pm 1 \pmod{29}
$$
and you may verify that both of these satisfy the original equation.  
Now, applying the CRT (calculating $[13^{-1}]_{29} = 9$ and $[29^{-1}]_{13} = 9$), we end up with $4$ solutions to the original equation (mod 377).  In particular, we have
$$
x \equiv 1 \pmod {377}\\
x \equiv -1 \pmod {377}\\
x \equiv 13[13^{-1}]_{29}(-1) + 29[29^{-1}]_{13}(1) \equiv 144 \pmod {377}\\
x \equiv 13[13^{-1}]_{29}(1) + 29[29^{-1}]_{13}(-1) \equiv -144 \pmod {377}\\
$$
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