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)=ax2+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,20162). How many values of n-mnm are possible?

1 Answer
Sep 15, 2016

165.165.

Explanation:

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.