# Let a,b,c,m and n be integers such that m<n and define the quadratic function f(x) = ax^2+bx+c where x is real. Then f(x) has a graph that contains the points (m,0) and (n, 2016^2). How many values of n-m are possible?

Sep 15, 2016

$165.$

#### Explanation:

f(x)=ax^2+bx+c, x in RR; a,b,c in ZZ

The graph of $f$ passes through pts. $\left(m , 0\right) , \mathmr{and} , \left(n , {2016}^{2}\right)$.

:. 0=am^2+bm+c....(1), &, 2016^2=an^2+bn+c.........(2).

$\left(2\right) - \left(1\right) \Rightarrow a \left({n}^{2} - {m}^{2}\right) + b \left(n - m\right) = {2016}^{2}$.

$\therefore \left(n - m\right) \left\{a \left(n + m\right) + b\right\} = {2016}^{2.}$

Here, $m , n , a , b , c \in \mathbb{Z} \text{ with } n > m$

$\Rightarrow \left(n - m\right) , \left\{a \left(n + m\right) + b\right\} \in {\mathbb{Z}}^{+}$

This means that $\left(n - m\right)$ is a factor of ${2016}^{2} = {2}^{10} \cdot {3}^{4} \cdot {7}^{2.} . . \left(\star\right)$

Therefore,

No. of possible values of $\left(n - m\right) ,$

$\text{=no. of possible factors of } {2016}^{2} ,$

$= \left(1 + 10\right) \left(1 + 4\right) \left(1 + 2\right) \ldots \ldots \ldots \ldots \ldots \left[b y , \left(\star\right)\right]$

$= 165.$

We have used this result : If the prime factorisation of $a \in \mathbb{N}$ is,

$a = {p}_{1}^{{\alpha}_{1}} \cdot {p}_{2}^{{\alpha}_{2}} \cdot {p}_{3}^{{\alpha}_{3}} \cdot \ldots \cdot {p}_{n}^{{\alpha}_{n}}$,

then $a$ has

$\left(1 + {\alpha}_{1}\right) \left(1 + {\alpha}_{2}\right) \left(1 + {\alpha}_{3}\right) \ldots \left(1 + {\alpha}_{n}\right)$ factors.