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?

1 Answer
Sep 15, 2016

Answer:

#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.