Of divisors and squares

* Edited solution. Thanks to Kushagra Sahni for his input. *

For a positive integer $m$ to have exactly four positive divisors we must either have that (i) $m = p^{3}$ for some prime $p$ or (ii) $m = pq$ for distinct primes $p,q.$

In case (i), since $m \le 120$ the only possibilities to consider are for $p = 2$ and $p = 3,$ since $5^{3} = 125 \gt 120.$ For $m = 2^{3} = 8$ the divisor sum is $1 + 2 + 4 + 8 = 15,$ and for $m = 3^{3} = 27$ the divisor sum is $1 + 3 + 9 + 27 = 40.$ Since neither of these divisor sums is a perfect square, this case yields no solution values for $m.$

In case (ii), for the sum of the divisors to be a perfect square, we require that

$1 + p + q + pq = n^{2} \Longrightarrow (p + 1)(q + 1) = n^{2}$ for some (positive) integer $n.$

Now as $p,q$ must be distinct we can assume without loss of generality that $p \lt q.$ Then since $11^{2} = 121 \gt 120$ we must have that $p \le 7,$ for otherwise $m = pq$ will exceed $120.$ Thus we only need look at $2,3,5,7$ as possible values for $p.$

With $p = 2$ we see that $q \le 59$ and that $3(q + 1) = n^{2} \Longrightarrow q + 1 = 3a^{2} \le 60$ for some integer $a \gt 1.$ As $3*5^{2} = 75 \gt 60$ the maximum value for $a$ is then $4.$ So checking the three possible remaining values for $a,$ we find that $(a,q) = (2,11), (3,26), (4,47),$ where in only the first and last of these pairings is $q$ prime. Thus $(p,q) = (2,11), (2,47) \Longrightarrow 2*11 = 22, 2*47 = 94$ are solution values for $m.$

With $p = 3$ we see that $q \le 37$ and that $4(q + 1) = n^{2} \Longrightarrow q + 1 = a^{2}$ for some integer $a.$ But $q = a^{2} - 1 = (a - 1)(a + 1)$ implies that $a = 2$ in order for $q$ to be prime, which in turn implies that $q = 3,$ which is not distinct from $p.$ Hence there are no solutions for $m$ with $p = 3.$

With $p = 5$ we see that $q \le 23$ and that $6(q + 1) = n^{2} \Longrightarrow q + 1 = 6a^{2}$ for some integer $a \gt 1.$ As $6*3^{2} = 54 \gt 23$ the only remaining possible value for $a$ is $2,$ which yields that $q + 1 = 6*2^{2} \Longrightarrow q = 23,$ which is indeed prime. Thus $m = pq = 5*23 = 115$ is another solution value.

Finally, with $p = 7$ we have that $q \le 17$ and that $8(q + 1) = n^{2} \Longrightarrow q + 1 = 2a^{2}$ for some integer $a \gt 1.$ The possible pairings for $(a,q)$ are then $(2,7)$ and $(3,17),$ with only the latter providing a distinct prime value for $q.$ Thus $m = pq = 7*17 = 119$ is our last solution.

The sum of the desired values for $m$ is then $22 + 94 + 115 + 119 = \boxed{350}.$