Show that the congruence equation $x^8 \equiv 16 \pmod p$ has a solution for every prime $p$. Show that the congruence equation  $x^8 \equiv 16 \pmod p$ has a solution for every prime $p$.
Let $g$ be a primitive root， then the equation is equivalent to $8 
\ ind_g(x)\equiv 4 \ ind_g(2) \pmod {p-1}$. It suffices to show that $2\mid ind_g(2)$.
If $p=2$, take $x=0$.
If $p\equiv \pm 1 \pmod 8$ we have $\left(\frac{2}{p}\right) = 1$, then there exists $a$ such that $a^2 \equiv 2 \pmod p$, so $a^8 \equiv 2^4 = 16 \pmod p$.
If $p\equiv 3 \pmod 8$ we have $\left(\frac{-2}{p}\right) = 1$, then there exists $a$ such that $a^2 \equiv -2 \pmod p$, so $a^8 \equiv (-2)^4 = 16 \pmod p$.
If $p\equiv 5 \pmod 8$ then $p\equiv 1 \pmod 4$, then there exists $a$ such that $a^2 \equiv -1 \pmod p$, so $(1+a)^2 \equiv 2a \pmod p$, then $(1+a)^4\equiv 4a^2 \equiv -4 \pmod p$ so $(1+a)^8 \equiv 16 \pmod p$.
If $p=2$, see jjagmath's answer.  $x^8-16=(x^2-2)(x^2+2)(x^2+2x+2)(x^2-2x+2)=0$
must have solutions mod $p$ (an odd prime),
because if $x^2-2$ and $x^2+2$ do not have roots mod $p$, then $x^2-\dfrac2{-2}=x^2+1$ does,
because  the inverse of a quadratic nonresidue is a nonresidue
and the product of two nonresidues is a residue.
Finally, if $a^2+1\equiv0\bmod p$, then $-1\pm a$ are roots of $x^2+2x+2$,
and $1\pm a$ are roots of $x^2-2x+2$, which would be obtained with the quadratic formula.
Thus, either $x^2-2$ or $x^2+2$ has roots mod $p$, or $x^2+2x+2$ and $x^2-2x+2$ have roots mod $p$;
in any event, $x^8-16$ has roots mod $p$.