I'm trying to understand why if the $gcd(p-1, 15) = d \neq 15$, then there are zero roots (since if it's $=15$, there are exactly 8). I was thinking that since a solution to $x^d - 1$ is relevant if $gcd(p-1, 15) = d \neq 15$, wouldn't the roots of $\Phi_{15}[x]$ technically still appear in $\mathbb F_p$ for $p = 1, 3, 5$? Why is there no in between and simple 8 solutions or none?

  • $\begingroup$ See your previous question. $\endgroup$ Mar 17, 2022 at 15:21
  • $\begingroup$ yes, this part I am still confused about in the discussion. I thought if $gcd = 1$ then there is one root since $x-1$ exists as a factor? I agree none exist for $p = 3,5$ $\endgroup$
    – webmathex
    Mar 17, 2022 at 15:25
  • $\begingroup$ Overthinking now. I guess panic induced. You have written down $\Phi_{15}(x)$ in the older question. If $x-1$ were a factor you would have $\Phi_{15}(1)=0$. It is not difficult to check by hand that this is not the case :-) $\endgroup$ Mar 18, 2022 at 13:05


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