, 2012

Since $p$ is an odd prime integer, then $\text{p gt;2}$.

Since the only prime divisors of $2p^2$ are 2 and $p$, then the positive divisors of $2p^2$ are $1,2,p,2p,p^2$, and $2p^2$.

So then, $S(2p^2)=1+2+p+2p+p^2+2p^2=3p^2+3p+3$.

Since $S(2p^2)=2613$, then $3p^2+3p+3=2613$ or $3p^2+3p-2610=0$ or $p^2+p-870=0$.

Factoring, $(p+30)(p-29)=0$, and so $p=29$ ($p\neq -30$ since $p$ is an odd prime).
Suppose $m=2p$ and $n=9q$ for some prime numbers $\text{p,q gt;3}$.

The positive divisors of $2p$, thus $m$, are $1,2,p$, and $2p$ (since $\text{p gt;3}$).

Therefore, $S(m)=1+2+p+2p=3p+3$.

The positive divisors of $9q$, thus $n$, are $1,3,q,3q,9$, and $9q$ (since $\text{q gt;3}$).

Therefore, $S(n)=1+3+9+q+3q+9q=13q+13$.

Since $S(m)=S(n)$, then $3p+3=13q+13$ or $3p-13q=10$.

Also, $m$ and $n$ are consecutive integers and so either $m-n=1$ or $n-m=1$.

If $m-n=1$, then $2p-9q=1$.

We solve the following system of two equations and two unknowns. $$\begin{align*} 2p-9q & = 1 \tag{1}\\ 3p-13q & = 10 \tag{2} \end{align*}$$ Multiplying equation (1) by 3 and equation (2) by 2 we get, $$\begin{align*} 6p-27q& = 3 \tag{3}\\ 6p-26q & = 20 \tag{4}\end{align*}$$ Subtracting equation $(3)$ from equation $(4)$, we get $q=17$.

Substituting $q=17$ into equation $(1)$, $2p-9(17)=1$ or $2p=154$, and so $p=77$.

However, $p$ must be a prime and thus $p\neq 77$.

There is no solution when $m-n=1$.

If $n-m=1$, then $9q-2p=1$.

We solve the following system of two equations and two unknowns. $$\begin{align*} 9q-2p & = 1 \tag{5}\\ 3p-13q & = 10 \tag{6} \end{align*}$$ Multiplying equation $(5)$ by 3 and equation $(6)$ by 2 we get, $$\begin{align*} 27q-6p & = 3 \tag{7} \\ 6p-26q & = 20 \tag{8}\end{align*}$$ Adding equation $(7)$ and equation $(8)$, we get $q=23$.

Substituting $q=23$ into equation $(6)$, $3p-13(23)=10$ or $3p=309$, and so $p=103$.

Since $q=23$ and $p=103$ are prime integers greater than 3, then $m=2(103)=206$ and $n=9(23)=207$ are the only pair of consecutive integers satisfying the given properties.
Since the only prime divisors of $p^3q$ are $p$ and $q$, then the positive divisors of $p^3q$, are $1,p,q, pq, p^2, p^2q, p^3$, and $p^3q$ (since $p$ and $q$ are distinct primes).

Therefore, $S(p^3q)=p^3q+p^3+p^2q+p^2+pq+p+q+1$.

Simplifying,

$$\begin{aligned} S(p^3q)&=p^3q+p^3+p^2q+p^2+pq+p+q+1\\ &=(p^3q+p^2q+pq+q)+(p^3+p^2+p+1)\\&=q(p^3+p^2+p+1)+(p^3+p^2+p+1)\\& =(q+1)(p^3+p^2+p+1)\\ &=(q+1)(p^2(p+1)+(p+1))\\ &=(q+1)(p+1)(p^2+1) \end{aligned}$$

We are to determine the number of pairs of distinct primes $p$ and $q$, each less than 30, such that $(q+1)(p+1)(p^2+1)$ is not divisible by 24.

There are 10 primes less than 30. These are $2,3,5,7,11,13,17,19,23$ and $29$.

Therefore, the total number of possible pairs $(p,q)$, where $p\neq q$, is $10\times 9 = 90$.

We will count the number of pairs $(p,q)$ for which $(q+1)(p+1)(p^2+1)$ is divisible by 24 and then subtract this total from 90.

If $p$ or $q$ equals 23, then 24 divides $(q+1)(p+1)(p^2+1)$.

There are 9 ordered pairs of the form $(23,q)$ and 9 of the form $(p,23)$.

Thus, we count 18 pairs and since we have exhausted all possibilities using 23, we remove it from our list of 10 primes above.

Since $24=2^3\times 3$, we can determine values of $q$ for a given value of $p$ by recognizing that each of these prime factors (three 2s and one 3) must occur in the prime factorization of $(q+1)(p+1)(p^2+1)$.

For example if $p=2$, then $(q+1)(p+1)(p^2+1)=(q+1)(3)(5)$.

Therefore, for $(q+1)(p+1)(p^2+1)$ to be a multiple of 24, $q+1$ must be a multiple of 8 (since we are missing $2^3$).

Thus when $p=2$, the only possible value of $q$ is 7 (we get this by trying the other 8 values in the list of primes).

We organize all possibilities for $p$ (and the resulting values of $q$) in the table below.

The total number of pairs $(p,q)$ for which 24 divides $S(p^3q)$ is $$18+1+5+7+5+8+4+7+5+7=67.$$ Thus, the total number of pairs of distinct prime integers $p$ and $q$, each less than 30, such that $S(p^3q)$ is not divisible by 24, is $90-67=23$.