Cool primes!

Claim: If $a^7 \equiv b^7 \pmod{p}$ , then $a \equiv b \pmod{p}$ .

First, if $b \equiv 0 \pmod p,$ then $a \equiv 0 \pmod p,$ which works. Then if $b \not\equiv 0,$ we can write $(ab^{-1})^7 \equiv 1 \pmod p.$ Then, the order of $ab^{-1}$ mod $p$ is either $7$ or $1,$ since the order must divide $7.$ But the order can't be $7,$ since $\phi(p) = 7k+2$ must be a multiple of the order, so it must be $1,$ implying that $a \equiv b \pmod p,$ as desired.

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Now, consider the set $\{0^7, 1^7, 2^7, \cdots , (p-1)^7 \}$ . Our claim shows that each of these numbers leave a distinct remainder modulo $p$ . Thus, modulo $p$ , this set is equivalent to $\{0, 1, 2, 3, \cdots , p-1 \}$ . We see that each of these numbers appear exactly once, i.e for all $k$ , there exists a $x$ such that $x^7-k$ is a multiple of $p$ . Further, note that the minimum possible value of $p$ is $3$ , which isn't cool as $3^7-3 \equiv 0 \pmod{3}$ . Our previous logic shows that no larger primes are cool. In conclusion, there exist no cool primes of the form $7k+3$ .