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*

*facepalm*

a_{2008}, obviously.

Maybe I'll do a few later, high school maths doesn't sounds especially appealing.

1)
Well let H_x be the number of y for each x.
H_x = H_-x

so the number we want is 2*Sum H_x

Now H_2008 = 1

Also H_n = H_(n+1) + 2, as any y satisfying n+1, also satisfies n, and then there are two more.

So H is a arithmetic sequence, apply the obvious formula, yada yada.
Most of the shit I've written is pretty obvious as well, the other thread was more fun :(

2) Induction almost certainly does it, might have to show P_n => P_n+2 if it's only true for even numbers in general, can't be arsed to check.


4) Well, P(1) = 1, P(2) = 2, P(5) = 5, P(26) = 26 etc.

But this means that if we take any n, and consider all the polynomials of degree n that satisfy this equation, we can find a sequence of n+1 numbers s.t P(a_n) = a_n for them all.

But two distinct polynomials of degree <= n cannot agree on more than n numbers, so the only possible polynomial is the identity.

5) no idea what inradius means.

6) Hmm, I'll think on it, this one looks more interesting

>>3
Yeah, these are mostly pretty simple I guess. :/

#1's just a pain in the ass, but I put it on there just to be list 'em all.

#2 I don't think induction works, at least not very simply, but there is a super easy way to do it.

#4 Yep.

#5 radius of inscribed circle.

#6 The only one remotely interesting, IMO.

I'll make the next set harder.

>>4
1) 4*(some triangular number [not on axis] + 2007 [on axis]) + 1 [(0,0)]

2) Start with obvious sequence: a_n = n.  That sum thing is 0, and hence even.  Take any arbitrary sequence.  Switch any 2 elements in the sequence.  Oh look!  Parity is preserved.  Doing enough switches, we can go from the obvious sequence to any other sequence.

>>5
#2 - yep.

The simplest way imo is to say that |a-b| is odd iff (a-b) is. So \Sigma |a_i-i| has the same parity as \Sigma(a_i-i) = 0.

Looks like the only one that isn't basically done is #6 (Spoiler: If you search for inradius on wikipedia, the second equation solves #5 instantly).

C'mon gaiz.

>>6
Always come up with the more elegant solution... I phail yet again.

>>5

Ah, nice. Should've thought of transpositions.

I'm still nowhere on 6

Not >>3

Can 6) be done using the chinese remainder theorem?

>>10
I dunno lol.  Maybe, but I don't see how.  That's not how I did it anyway.

>>1 here, and also the guy from the other thread.

I give up on 6. Someone do it cause it's bugging me, I have no idea where to even start.

>>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.

>>13
Ah, that makes more sense.

I was going to go to the trouble of trying to construct a counterexample for the original question, which wouldn't have be too hard I don't think, seemed to me letting Q = x - 232 and then for almost any p you choose, there's almost certainly a solution to P(x) = 231, just as long as the leading co-efficient is positive.

>>14
Well in any case P(Q(x))=1 is a polynomial equation, and so there is always a complex solution.

If you wanted a real solution, all you have to do is rig things up so that the degree of P(Q(x)) is odd.  If P(x) = x^2(x-1) and Q(x) = x - 232, then P(Q(x))-1=(x-232)^2(x-233)-1=0 has odd degree and so a real root.

For a rational solution...hmmm I'm not sure.

>>3
>Maybe I'll do a few later, high school maths doesn't sounds especially appealing.
>high school math doesn't sounds

Good thing this isn't the English board.

>>16
Oh, hey, good one.

>>16
There's an english board?  Isn't every board an english board?

>>18
In that case every board would be a mathematics board but most people aren't aware of the calculations their brain has to make to type a few English words.