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?

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Answer 1 · 2016-09-15T15:00:27

$165.$

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

The graph of $f$ passes through pts. $(m,0),and, (n,2016^2)$ .

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

$(2)-(1) rArr a(n^2-m^2)+b(n-m)=2016^2$ .

$:. (n-m){a(n+m)+b}=2016^2.$

Here, $m,n,a,b,c in ZZ" with "n>m$

$rArr (n-m), {a(n+m)+b} in ZZ^+$

This means that $(n-m)$ is a factor of $2016^2=2^10*3^4*7^2...(star)$

Therefore,

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

$"=no. of possible factors of "2016^2,$

$=(1+10)(1+4)(1+2)...............[by, (star)]$

$=165.$

We have used this result : If the prime factorisation of $a in NN$ is,

$a=p_1^(alpha_1)*p_2^(alpha_2)*p_3^(alpha_3)*...*p_n^(alpha_n)$ ,

then $a$ has

$(1+alpha_1)(1+alpha_2)(1+alpha_3)...(1+alpha_n)$ factors.

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