Solve the equation in the field

Question

How to solve the $1+x+x^2+x^3+x^4+x^5+x^6=0$ equation in the $F_{29}$?

I've found that $(a+b)^{p}=a^{p}+b^{p}$ - perhaps it might help.

Answer 1

I am not sure if your methods will work. By the way, the formula for $a^p+b^p$ works because the coefficients are the multiples of $p=29$. Try to write some out and you will see.

My method

First multiply your polynomial by $(x-1)$ to get $x^7-1$. Then try to solve the latter. Since we add a factor $(x-1)$, all we have to do is to collect the solutions to the latter, and get rid of $1$ (you may check $1$ is not a solution).

The main trick is to notice that $F_{29}^* := F_{29} -\{0\}$ is a group (of order $28$) under multiplication! So every nonzero element $a$ must satisfy $a^{28}=1$. However, $28 = 4\cdot7$, so you can easily produce lots of roots by looking at the fourth powers of the nontrivial elements.

Answer 2

$(x-1)(x^6+\cdots+x+1)=x^7-1$, so the roots of your polynomial are the non-trivial seventh roots of unity. As $29-1=7\times 4$, the seventh roots of unity in $GF(29)$ are the fourth powers of the nonzero elements, and it's easy to find those.