Showing $x^8\equiv 16 \pmod{p}$ is solvable for all primes $p$

One way is to use the Legendre symbol identity $2^{(p-1)/2} \equiv (\frac{2}{p}) \equiv (-1)^{(p^2-1)/8} \pmod p$ (for odd primes p), keeping in mind that if $(8,p-1)=8$ then $p \equiv 1 \pmod 8$.

HINT $\rm\ \ \ x^8 - 16\ =\ (x^2 - 2)\: (x^2 + 2)\: (x^4 + 4)\:.\ \:$ If the first two factors have no roots in $\rm\ \mathbb Z/p\ $ then $\:2, -2\:$ are nonsquares so their product $-4\: $ is a square, so $\rm\: i = \sqrt{-1} \in \mathbb Z/p\:$. Thus the third factor has a root since $\rm\ x^4 + 4\ $ has roots $\rm\: \pm 1\pm i\:$.

I usually set this as an exercise when teaching Number Theory. My hint is to ask the students: what are the solutions of $z^8=16$ in the complex numbers?