An interesting discovery

Suppose that ${a}$ and ${b}$ are both positive integers. Now we must find the gradient and intercept of a line/chord joining (- ${a}$ . $\ a^{2}$ ) and ( ${b}$ , $\ b^{2}$ ). Using: $\ y-y_{1} = m(x-x_{1})$ : $m = \frac{b^2 - a^2}{b+a} = \frac{(b+a)(b-a)}{b+a} = b-a \Rightarrow\ y - b^2 = (b-a)(x-b)$ $\therefore\ y = (b-a)x +ab \ \ with \ ab = 120$ Now the number of factors can be calculated as follows ( $\ p_{i}$ are all primes ): $\psi (p_1^{a_1} p_2^{a_2} ...p_n^{a_n}) = (a_1 +1)(a_2 +1)...(a_n +1)$ $\Rightarrow\psi (120) = \psi(2^3 \times\ 3 \times\ 5 ) = 4 \times\ 2 \times\ 2 = 16$ Now we are asked to look at the different possible gradients so it is important to note that m= (b-a) means that (a,b) = (x,y) and (a,b) = (-y, -x) will produce the same gradient so we ignore all negative factors of 120.

$\therefore\ answer = 16$