Dealing with primes

Let $4 + pq = m^{2}$ and $4 + pr = n^{2}$ for some positive integers $m,n$ . Then

$pq = m^{2} - 4 = (m - 2)(m + 2)$ and $pr = n^{2} - 4 = (n - 2)(n + 2)$ .

Now we cannot have $p = q = 2$ , since then $4 + pq = 8$ would not be a perfect square. Similarly for $p = r = 2$ . So the minimum value for either of $4 + pq$ or $4 + pr$ must be $4 + 2 \times 3 = 10 \gt 3^{2}$ , and thus $m,n \gt 3 \Longrightarrow (m - 2), (n - 2) \gt 1$ . Thus by the Fundamental Theorem of Arithmetic,

(i) either $p = m - 2, q = m + 2$ or $p = m + 2, q = m - 2$ , and

(ii) either $p = n - 2, r = n + 2$ or $p = n + 2, r = n - 2$ .

Now if $p$ were to equal both $m - 2$ and $n - 2$ then $m = n$ , which would imply that $q = r$ , which is not allowed. So without loss of generality let $p = m + 2 = n - 2$ . Then $q = m - 2 = p - 4$ and $r = n + 2 = p + 4$ . So we are looking for sequences of primes of the form $p - 4, p, p + 4$ .

Now we cannot have $p - 4 = 2$ as this would make $p = 6$ , which is not prime.

If $p - 4 = 3$ then we would have the valid prime sequence $3, 7, 11$ .

If $p - 4 \gt 3$ then as $p - 4$ must be prime we must have either

$p - 4 \equiv 1 \pmod{3} \Longrightarrow p \equiv 2 \pmod{3} \Longrightarrow p + 4 \equiv 0 \pmod{3}$ , or

$p - 4 \equiv 2 \pmod{3} \Longrightarrow p \equiv 0 \pmod{3} \Longrightarrow p + 4 \equiv 1 \pmod{3}$ .

In either case one of the next elements of the sequence will be divisible by $3$ and hence not prime. So the only possible sequence of primes is $3,7,11$ . Thus $n = 1$ , and the desired solution is $1 + 3 + 7 + 11 = \boxed{22}$ .