What numbers are not in this set?

We will construct all elements of $S$ :

We might observe that if $n\in S$ , then $n+5\in S$ , since $(n+5)^2\in S$ . We can extend this to the fact that $n+5k, (n+5k)^2\in S$ for positive integers $k$ (call this "statement (A)", it will be useful later).

So, since $2\in S$ , we know that $x\in S$ if $x \equiv 2 \mod 5$ . We will now show that $3, 4, 11\in S$ :

Note that, by statement (A), that $49^2\in S$ =2401, thereby implying that $2401 + 5k\in S$ or, if $k = 2c$ , $2401 + 10c\in S$ and therefore, $3^8 = 6561\in S$ . We can see that $3^8 = 3^4)^2$ , and so $3^4\in S$ . We can repeat this to find that $3^2$ and $3\in S$ .

Since $3^4=81\in S$ , we also have that $121\in S$ by statement (A), and therefore $11\in S$ .

A similar argument shows that $4 \in S$ :

$11 \in S$ implies that $(11+5)^2 = 16^2 = 4^4 \in S$ , and so $4^2$ and $4 \in S$ .

We can show that $1\not\in S$ by noting that since $S$ is minimal, all elements of $S$ can be constructed by invoking the given rules. Clearly, only the second rule would allow us to show that $1 \in S$ , but that would imply that $-6$ or $-4 \in S$ , contradicting the positivity of $S$ .

We can also show that no integer $f$ satisfying $f \equiv 0 \mod 5$ :

Note that by statement (A) and the arguments resented above, if any multiple of 5 is in $S$ , then all multiples of 5 are in $S$ . However, for any multiple of 5 to be in $S$ , we must already have an element of $S$ whose square is a multiple of 5. However, we do not have this, since if $x \equiv 0 \mod 5$ , then $x^2 \equiv 0 \mod 5$ . So, there are no multiples of 5 in $S$ .

Therefore, $S$ is equivalent to the set of positive integers $\{x > 1: x \not\equiv 0 \mod 5\}$ . Therefore, $1$ , $5$ , and $10$ are not in $S$ , so the answer is $1+16+512 = 529$ .