Fifth power remainders

FIrstly, if $5 \nmid p - 1$ , then $\gcd(5, p - 1) = 1$ . We claim that $a^5 \equiv b^5 \pmod{p} \Rightarrow a \equiv b \mod{p}$ . The case when $p \mid a$ is obvious, so assume that $a$ and $b$ are relatively prime to $p$ . Note that by Fermat's little theorem, $a^{p - 1} \equiv b^{p - 1} \equiv 1 \pmod{p}$ . So, $a^{\gcd(5, p - 1)} \equiv b^{\gcd(5, p - 1)} \pmod{p}$ , which implies $a \equiv b \pmod{p}$ . Thus, by the claim, there exists an integer $n$ such that $n^5 \equiv 2 \pmod{p}$ . (For { $1^5, 2^5, ..., (p - 1)^5$ } = { $1, 2, ..., p - 1$ } mod $p$ ). This implies $f(p) = 2$ . So, the minimum possible value of the fraction $\frac{p}{f(p)}$ purely depends on $p$ . Thus, the minimum value of the fraction for this case is $\frac{13}{2}$ . (Since $p > 7$ .)

Secondly, let $5 \mid p - 1$ . Then, $p = 10k + 1$ for some integer $k$ . By direct computation for the cases $k = 1, 3, 4, 6, 7$ and $10$ , we find that the two smallest possible values of the fraction is $\frac{11}{10}$ and $\frac{71}{20}$ (obtained from $k = 1, 7$ .) . It is clear that $\frac{71}{20} < \frac{13}{2}$ , so we just have to check that there are no more smaller fractions for $k > 10$ . Since for $k = 11$ , $p$ is not a prime, we just have to check for $k \ge 12$ or $p \ge 121$ . For $p \ge 121$ , $2^5 \equiv 32 \pmod{p}$ . So, $f(p) \le 32$ . Thus, $\frac{p}{f(p)} \ge \frac{p}{32} \ge \frac{121}{32} \ge \frac{71}{20}$ . This implies that the two smallest possible value of the fraction is indeed $\frac{11}{10}$ and $\frac{71}{20}$ . Their sum is $\frac{93}{20}$ . So, $a + b = \boxed{113}$ .

Note : $f(11) = 10, f(31) = 5, f(41) = 3$ , $f(61) = 11, f(71) = 20 , f(101) = 6$ .

If p-1 is relatively prime to 5, f(p) is always 2, so the best we can do is 13/2.

Now suppose p-1 is divisible by 5, i.e. p = 10k + 1. Direct computation for k = 1, 3, 4, 6, 7, and 10 gives that the best two fractions are 11/10 and 71/20, which are both less than 13/2. There is also no need to continue further, since 2^5 is always a fifth power and 71/20 < 131/32 <= p/f(p) for p > 101.