$k$ is not perfect though

We are looking for positive integer $k$ such that $k^2 + 15k = p^2$ where p is any positive integer. Let's try converting this to standard form: $k^2 + 15k - p^2 = 0$ What we see above is a quadratic equation with $a = 1, b = 15, c = -p^2$ , let's consider the formula for the roots (i.e. the value of k). $k = \frac{-b \pm \sqrt{ 15^2+4p^2}}{2}$ . Let $d = 15^2+4p^2$ .

Since we know k is a integral, $-15+\sqrt(d)$ should be divisible by 2. That only happens when d is a perfect square and is odd. So, we want to determine if there exists a $q$ such that $15^2 + 4p^2 = q^2$ where p and q integers and p < q.

The difference between two perfect squares $q^2$ and $p^2$ is $2 kp + k^2$ where $k = q - p$ . Hence, we need to check if there exists a k (i.e. p - q) such that $(15^2 + 4p^2) - p^2 = 2kp + k ^2$ .

The equation when simplified is equivalent to the quadratic two-variable Diophantine equation. $3p^2 - 2kp - k^2 + 15^2 = 0$

I don't actually know how to manually solve for the solution of this type of equation so I just used a solver: https://www.alpertron.com.ar/QUAD.HTM It will turn out the equation has 4 solutions with p > 0 and q > 0. Use the quadratic formula to get the k's for each p. You will get k = 1, 5, 12, 49 which sum to 67.

I will edit this solution later