You should be able to solve this

Round 2, bitches.

These are from a state high school math competition that was held at my uni a while ago.  As far as I know, the problems aren't anywhere on the interbutts.

1)  How many pairs of integers (x,y) satisfy |x|+|y| \le 2008? (Answer should be an integer)

2) Take the integers 1, 2, \ldots ,2008 and write them down in some order to give the sequence a_1, a_2, \ldots, a_2008.  Prove that

|a_1-1| + |a_2-2| + \ldots + |a_2008-2008|

is even.

3) (Some geometry question with a diagram and crap. Forget it.)

4) Find, with proof, all polynomials P(x) such that P(0) = 0 and, for all integers x,

P(x^2+1) = \left( P(x) \right)^2 + 1.

5) A triangle has area 2008 and perimeter 1492.  What is its inradius?

6) P(x) and Q(x) are polynomials with integer coefficients.  There is some integer a such that both a and a+1 are roots of P(x).  In addition you know that Q(2008) = 1776.  Prove that the equation Q\left(P(x)\right) = 1 has no solution.

*crosses fingers*

>>12
O SHI-

LMAO I just realized the problem was supposed to say "no INTEGER solutions"  >.> Sooooooo sorry!!  Here, I'll post the answer anyway just so we can end the thread and be done with it.

Since a, a+1 are roots of P(x), P(x) = (x-a)(x-a-1)F(x) for another integer polynomial F.  So if x is any integer, either x-a or x-a-1 is even, and P(x) is even.

Since Q(2008) = 1776, the constant term of Q(x) is even (write out the equation explicitly), and Q(x) is even whenever x is. Therefore, etc.