What are ALL the possible values for $n$ and show there are no more.

Question

I have been doing some practice questions for an upcoming Maths Challenge. There's one question I can't seem to grasp. I'm not sure entirely sure where to start. I don't know how to approach this one. Any help would be appreciated

$n$ in the form $n = pq$ where $p$ and $q$ are different odd prime numbers is not superdeficient. Not being superdeficient means the sum of $n$'s proper factors less than $n$, when multiplied by $2$, is more than $n$.

E.g. $35$ is superdeficient since $1+5+7=13 \times 2=26<35$ but $15$ isn't since $1+3+5=9 \times 2=18>15$.

Answer 1

We want to find all pairs of distinct odd primes $p < q$ such that $pq$ is not superdeficient i.e., it satisfies $1+p+q>\frac{pq}{2}$. Note first that if $p \geq 5$, then $\frac{pq}{2} \geq \frac{5}{2}q \geq 2q>1+p+q$, a contradiction. Hence $p=3$. Hence it remains to find primes $q>3$ which obey $4+q>\frac{3}{2}q$. These are only $q=7, q=5$

Concluding, we have solutions $(3,5)$ and $(3,7)$.

Answer 2

$p \ne q \implies \text{(WLOG) } p < q.$

$2(1+p+q) \ge pq\implies p \le \dfrac{2(1+q)}{q-2}.$

$q = 5 \implies p = 3 \implies n = 15.$

$q = 7 \implies p = 3 \implies n = 21.$

At this point notice that it can't be $q \ge 11,$ since $\dfrac{2(1+11)}{11-2}=\dfrac{8}{3}<3$ and $\dfrac{2(1+x)}{x-2}$ is decreasing $\forall x \ge 2$.