# Number of roots in cyclotomic polynomial $\Phi_{15}[x]$ in $\mathbb F_p$

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?

• See your previous question. Mar 17, 2022 at 15:21
• 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 $p = 3,5$ Mar 17, 2022 at 15:25
• 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 :-) Mar 18, 2022 at 13:05