I have to find a prime p such that the 35th cyclotomic polynomial has an irreducible 12th-degree factor in $F_p$.
I suppose that the roots of the 35th cyclotomic polynomial have order 1,5,7 or 35. The only element with order one is the identity multiplicative element, 1. And $\phi _{35} (1) = 1$ not $0$, so we know that the roots have order 5,7 or 35.
The extension of $F_p$ to find the irreducible 12th-degree factor has $p^{12}-1$ elements.
By Lagrange, the order of any element of a field K has to divide the order of K$^×$.
So, I take for example an element of order 35. I think that I have to try with primes for search any prime that 35 divides at $p^{12}-1$. For example, $p=2$.
Would this reasoning work?