Sequence and Series question (maybe)?

[Echo8]

1. The problem statement, all variables and given/known data

Find all positive integers n such that both n + 2008 divides n^2 + 2008 and n + 2009 divides n^2 + 2009

2. Relevant equations

-

3. The attempt at a solution

I have no idea where to start.... I'm not even sure it's a sequence and series question. If it is then I have no idea what to do.
I've played around with it by saying (n^2+2008)/(n+2008)=a and (n^2+2009)/(n+2009)=b, rearranging them into quadratics and equating them, then rearranging again in terms of n. Doing this I got n=(-2008a+2009b-1)/(a-b) and couldn't see anything to do from this. I did a similar thing using partial fractions but ended up with pretty much the same equation (though not exactly the same so I could well have done something wrong but I don't think that was the right way to go about it anyway).
Any light that anyone can shed on this would be much appreciated :)

[micromass]

One way that n+2008 could divide n^2+2008 is that n+2008=n^2+2008. But then n=0 or n=1.

Assume that n+2008 divides n^2+2008. Then n+2008 also divides (n^2+2008)-(n+2008). Thus n+2008 divides n^2-n.
Now do thesame for n+2009, what do you get?

[Echo8]

I get n+2009 divides n^2-n.
I see that they have the same numerator... Though dividing n+2008 and n+2009 to get rid of n^2 - n won't necessarily give another integer so I'm guessing that's not the next step. I had a play around after doing this and got n^2 - n = n^2 - n which also isn't much help...!
This is a little embarrassing :blushing:

[micromass]

Yes, and since gcd(n+2008,n+2009)=1. We must have that (n+2008)(n+2009) divides n^2-n. This is a contradiction.

[Echo8]

Yes I see it now, thanks for the help :)