Find all natural number $n$ such that $n^2 + 2\mid 2001n + 2$
My idea is to eliminate $n$ on the right side of divisibility sign so it becomes only an integer, without any variable. $$\label{eq1} n^2 + 2\mid 2001n + 2\tag{1} $$
Here's my approach. To eliminate $n$, we need another linear "right side". So we try to find something looks like $$n^2 + 2 \mid a(n^2 + 2) - b(2001n + 2)$$ where $a$ and $b$ are function of $n$. A suitable choice would be $a = 2001$ and $b = n$, which gives $$n^2 + 2 \mid 2001(n^2 + 2) - n(2001n + 2)$$ thus $$\label{eq2} n^2 + 2 \mid 4002 - 2n\tag{2} $$ Now we have two linear right sides, eliminating $n$ by multiplying \eqref{eq1} by $2$ and \eqref{eq2} by $2001$ gives us $$n^2 + 2\mid 2(2001n + 2) + 2001(4002 - 2n) = 4 + 2001\cdot 4002$$ or $n^2 + 2\mid 8008006$. We only need to find the divisor of $8008006$ for which when we subtract 2 from it will give a square number. But this is somewhat laborious. Is there any other way to solve this?
Edit: The solutions I got $n = 6,9,2001$
Thanks!
Well, we know $2001n+2 \geq n^2+2$. Then $2001n-n^2 | n^2+2$ or $n^2+2|2001n-n^2$. Assume the first case is true. We divide this into two subcases: $n$ even and $n$ odd. If $n$ odd, then we have a clear contradiction as the right side is $2 \pmod n$ and the left side is divisible by $n$. Otherwise, unless $n=2$, clearly $n$ must have a prime divisor other than 2 that divides the left side but not the right. Unfortunately for $n=2$, the left side is orders of magnitude larger than the right side. For the second case, we have the same issues for $n \neq 2$, and plugging in $n=2$ unfortunately leads us to conclude there are no solutions.
EDIT: It appears I forgot the case where $2001n-n^2=0$. This yields the solution of $n=2001$. Oops. There is also a huge gaping hole in the second case. My logic for the first case does NOT hold for the second case.