function

This is sort of easy.. but it should click!

Let g(x)= f(x)-2008
a) g has atleast 4 roots.
b) g is a polynomial! (since f is a polynomial)

g(m) =f(m) - 2008
= 2008-2008. = 0 (hece g(m) = 0)

hence m is a root of g.. similarly for n,o,p
hence these 4 are the roots of g!

we need to prove that
f(x) ≠2010 for any k
f(x) = g(x) + 2008
so, g(x)+2008≠2010 for any k
so we need to prove g(x) ≠2 for any k !

g(x) = (x-m)(x-n)(x-o)(x-p).t(x) .. where t is a polynomial!
since m,n,o,p are roots of g!

read now.. again.. (I have editied it :)

by that time i will post the remaining solution

g(x) = (x-m)(x-n)(x-o)(x-p).t(x)

if we can prove that for any k, g(x) is never 2, we are done!

m,n,o,p are distinct.

for g(x) to be 2, it can have only factors in -1,1 or 2 or -2

but not both (2 and -2)

any other factor it will be not equal to 2..

but there are 4 integers (all will be uniqe! since m,n,o,,p are difft!)

x-m
x-n
x-o
x-p

we can also assume that

x-m<x-n<x-o<x-p (because we can take m,n,o,p in decreasing order!

So x-m=-1 (it is the least!)
x-n=1
x-o=2

what do we do of x-p???

same if we assume
x-m=-2
x-n=-1
x-0=1
x-p cannot be +2! (the product will be 4!)