No Other Prime Must Divide It

For $\omega(n)$ to be equal to $1$ , we need $n$ to be equal to $q^r$ where $q$ is any prime no. and $r \epsilon N$ .

Therefore, we need $(p-1)!+1 = q^r$

But, observe that -

$(p-1)!+1 \equiv 0 \pmod{p}$ (By Wilson Theorem)

$\Rightarrow (p-1)!+1 = pm$ (for some $m$ )

But, again remember that $(p-1)!+1 = q^r$ .

This suggests that $m$ should be equal to $p^{k-1}$ for some arbitrary $k$ which enables us to write

$(p-1)!+1 = p^k$

$\Rightarrow (p-1)! = p^k-1$

Now, we suspect that the solutions to the above equation are ${2,3,5}$ only.

This can easily be proved by considering the following inequality for $p \geq 7$

$(p-1)^2|(p-1)!$

$\Rightarrow (p-1)^2|p^k-1$

$\Rightarrow p-1|1+p+p^2+\ldots p^{k-1}$

$\Rightarrow p^k-1 > (p-1)!$

Thus, the only solutions are ${2,3,5}$ which sum up to $2+3+5 = \boxed{10}$

wilson and liouville

Sorry, I will post a better solution involving proper methods later. For now, I solved this question by hit and trial. We see that for $2$ , $3$ and $5$ ...the answers are $2$ , $3$ and $25$ respectively which have exactly one prime divisor (unique). Verifying for $7$ and $11$ , we find that they don't work. Thus answer= $2+3+5= {10}$