There is a problem I am trying to solve involving cyclotmic polynomials ( $\Phi_n(x)$), that any answer, if found, will help me apply some knowledge to field extensions of cyclotomic fields.
Suppose we have a prime number $p = 1 \pmod n$, or a number $p$ such that each prime $q$ dividing $p$ is also congruent to $1 \pmod n$. Is it easy to solve for all values of $x$ in $\Phi_n(x) = 0 \pmod p$, or namely find at least one value of $x$ which is a solution?
One method I do know if $p$ and $n$ are both prime is to find a solution $x > 1$, choose any base $b$, and compute $b^{(p-1)/n}$ $\pmod p$ $= x$. Then $x^n = 1 \pmod p$, and $x^n-1 = 0 \pmod p$, and since $\Phi_n(c)$ divides $x^n-1$, $p$ divides $\Phi_n(x)$, these solutions are applicable if $x > 1$.
For what I am trying to solve (for $x$), is
$\Phi_{53}(x) = 0 \pmod {20353*107^{102}}$
There are solutions $x$, because both prime factors $20353$ and $107$ are congruent to $1 \pmod {53}$. Using the method I described earlier with base $b$ does not work since,
$b^{40704*107^{101}}$ $\pmod {20353*107^{102}}$ seems to be $1$ for all $b$, but $1$ is a trivial solution, and I am looking for solutions $x > 1$.
Any help please? Thank you.