How do you compute $1+x+\cdots+x^7=0 \mod 40$?

Question

I think the first step is to us the fact that $(x-1)(x^7+\cdots+x+1)=x^8-1$.

This makes the problem into find $x^8=1 \mod 40$, which is where I am stuck. I know that $x$ can't be even or one, but I do not know a process that will find the actual values without brute force computations. I'm not sure how to know it is solvable, but because of Wolfram I know it is. I know $\phi(40)=16$, but not sure where that helps. Any help is appreciated, thanks in advance.

Answer 1

$\lambda(40)=4 \Rightarrow x^8 \equiv 1 \pmod{40}$ iff $\gcd(x,40)=1$