elementary number theory - Find all positive integers $n$ such that $n+2008$ divides $n^2 + 2008$ and $n+2009$ divides $n^2 + 2009$

I wrote

$$\begin{align} n^2 + 2008 &= (n+2008)^2 - 2 \cdot 2008n - 2008^2 + 2008 \\ &= (n+2008)^2 - 2 \cdot 2008(n+2008) + 2008^2 + 2008 \\ &= (n+2008)^2 - 2 \cdot 2008(n+2008) + 2008 \cdot 2009 \end{align}$$
to deduce that $n+2008$ divides $n^2 + 2008$ if and only if $n+2008$ divides $2008 \cdot 2009$.

Similarly, I found that $n+2009$ divides $n^2 + 2009$ if and only if $n+2009$ divides $2009 \cdot 2010$.

I can't seem to get anywhere from here. I know that $n=1$ is an obvious solution, and I think it's the only one, although I can't prove/disprove this. I know that any two consecutive integers are coprime but I haven't been able to use this fact towards the solution.

Could anyone please give me any hints?

For the record, this is a question from Round 1 of the British Mathematical Olympiad.

Answer

Here another approach

Since $(n+2008)|( n^{2} + 2008)$ and $(n+2009) |( n^{2} + 2009)$ then

$(n+2008)k = n^2 + 2008$ and $(n+2009)m = n^2 + 2009$ for some $k,m \in \mathbb{Z}$

$\textbf{Case 1:}$ $k=1$ or $m=1$

$(n+2008) = n^2 + 2008$ or $(n+2009) = n^2 + 2009$

in any case leads to $n^2-n=0$ that is $n=1$ and $n=0$

$\textbf{Case 2:}$ $m>k>1$

$$n^2+2009=(n+2009)m = (n+2008)m+m > (n+2008)k + m > n^2+2009$$ which is impossible

$\textbf{Case 3:}$ $k>m>1$

$$n^2+2008=(n+2009-1)k = (n+2009)k - k \ge (n+2009)(m+1) - k = (n+2009)m +(n+2009-k) \ge n^2+2009$$ also impossible, the last inequality holds because $k< n+2009$, suppose not

$k \ge n+2009 >n+2008 \Rightarrow n^2 + 2008 =(n+2008)k > (n+2008)^2 > n^2 + 2008$ a contradiction

$\textbf{Case 4:}$ $k=m>1$

$$(n+2008)k+1=(n+2009)k \Longrightarrow k =1 \Longrightarrow n=1$$

So the solutions are $\boxed{n=1}$ and $\boxed{n=0}$