jmc

We know that $2019^8\equiv -1(\mathrm{mod}p)$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we find $2019^{16}\equiv 1(\mathrm{mod}p)$. Since $2019^{16}\equiv 1(\mathrm{mod}p)$, the order of $2019$ modulo $p$ is a positive divisor of $16$. However, if the order of $2019$ modulo $p$ is $1,2,4,$ or $8,$ then $2019^8$ will be equivalent to $1(\mathrm{mod}p),$ which contradicts the given requirement that $2019^8\equiv -1(\mathrm{mod}p)$. Therefore, the order of $2019$ modulo $p$ is $16$. Because all orders modulo $p$ divide $\varphi (p)$, we see that $\varphi (p)$ is a multiple of $16$. As $p$ is prime, $\varphi (p)=p\left(1-\frac{1}{p}\right)=p-1$. Therefore, $p\equiv 1(\mathrm{mod}16)$. The two smallest primes equivalent to $1(\mathrm{mod}16)$ are $17$ and $97$. As $2019^8\not\equiv -1(\mathrm{mod}17)$ and $2019^8\equiv -1(\mathrm{mod}97)$, the smallest possible $p$ is thus $\boxed{97}$.