Let $f(i)$ be the number of integer $t$ not exceeding $2017$ satisfying $P(t)\equiv i\pmod{7}$, and $S_i$ be the set of such integers. Claim: if $f(i)$ is odd, then a wolf would eat a sheep. Otherwise, no wolf dares to eat a sheep. We prove by induction. Base cases: If $f(i)=1$ then the only wolf in $S_i$ can eat sheep $i$ recklessly If $f(i)=2$ either one will be scare to be eaten by the other one and therefore will not eat the sheep. Inductive step: if $f(i)$ is odd, a wolf would eat a sheep knowing that after doing so, the case would become $f(i)$ is even for the rest of the wolves and would not dares to eat it. if $f(i)$ is even, if a wolf eats a sheep, he case would become $f(i)$ is odd for the rest of the wolves and it would got eaten recklessly. So any wolf would not dares to eat a sheep. Now note that $1, 2$ are in $S_7$, $3$ is in $S_1$, $4, 5$ is in $S_2$, and so on. As we know that the difference of two consecutive primes is even except for $2$ and $3$, and with the fact that $2017$ is a prime, we know that only $S_1$ has an odd number of element, so only $f(1)$ is odd. Therefore, only one sheep would be eaten once, giving us $2016$ wolves.