Douglass's Sum

First of all, if $\frac{q^2+15q+42}{p}$ is an integer, then the denominator must be divisible by the numerator i.e. $p\mid (q^2+15q+42)$ . If $p\mid q$ , then we have $p\mid 42$ . Hence, $p$ can be any positive divisor of $42$ . $42$ has $8$ positive divisors which are $1,2,3,6,7,14,21,42$ . And the sum is $\fbox{96}$ .

Note $\frac{q^{2}+15 q + 42}{p} = \frac{q^{2}+15q}{p} + \frac{42}{p}$ . This says that $p \mid 42$ Now its easy to see that $\sigma(42) = 96$

p divides q therefore q^2 and 15q. Hence for the given value to be an integer p should divide 42. So the values of p are 1,2,3,6,7,21,14 and 42. sum of these values of p is equal to 96.

$p > 0,\: p|q$ so clearly if $p | \left(q^2 + 15q + 42\right)$ then $p|42$ and thus $p \in \{1,2,3,6,7,14,21,42\}$

Thus, the sum of all $p$ is $1+2+3+6+7+14+21+42 = 96$

Since $p$ $\vert$ $q$ , $q$ can be written as $pk$ , where $k$ $\epsilon$ $\mathbb{Z}$

The fraction $\frac{q^{2} + 15q + 42}{p}$ can be written as

$\frac{q^{2}}{p} + \frac{15q}{p} + \frac{42}{p}$

$= \frac{p^{2}k^{2}}{p} + \frac{15pk}{p} + \frac{42}{p}$

$= pk^{2} + 15k + \frac{42}{p}$

Since $p$ and $k$ are integers, for the above expression to be an integer, $\frac{42}{p}$ must also be an integer.

That is, $p$ $\vert$ $42$ .

So $p$ has to be a factor of 42.

Factors of $42$ are $\{1, 2, 3, 6, 7, 14, 21, 42\}$

The answer to this problem is $1 + 2 + 3 + 6 + 7 + 14 + 21 + 42$

= $\boxed{96}$

First, we say $q=ap$ , since $p$ divides $q$ . Next, we divide the fraction up into $\frac{(ap)^2}{p} +\frac{15ap}{p}+\frac{42}{p}$ . The first two terms are integers, so we must find the values of $p$ where $\frac{42}{p}$ are integers, a.k.a the divisors of $42$ . These are ${1,2,3,6,7,14,21,42}$ , which have a sum of $\fbox{96}$