<s>[INST] positive Integer value of $n$ for which $2005$ divides $n^2+n+1$ 
How Can I calculate positive Integer value of $n$ for which $2005$
  divides $n^2+n+1$

My try:: $2005 = 5 \times 401$
means $n^2+n+1$ must be a multiple of $5$ or multiple of $401$
because $2005 = 5 \times 401$
now $n^2+n+1 = n(n+1)+1$
now $n(n+1)+1$ contain last digit $1$ or $3$ or $7$
$\bullet $ if last digit of $n(n+1)+1$ not contain $5$. So it is not divisible by $5$
Now how can I calculate it? please explain it to me.
 [/INST] If a number is divisible by $2005=5\cdot401,$ it must be divisible by $5$ and by $401$
Now, as you have identified $n^2+n+1$ is not divisible by $5,$ it can not be divisible by  any multiple of $5$ like  $2005=5\cdot401$

Alternatively,
$$n^2+n+1\equiv0\pmod {2005}\implies n^2+n+1\equiv0\pmod 5$$
$$\implies 4(n^2+n+1)\equiv0\pmod 5\implies (2n+1)^2\equiv-3\equiv2$$
Now, $(\pm1)^2\equiv1\pmod 5,(\pm2)^2\equiv4$
So, there is no solution to $ (2n+1)^2\equiv-3\pmod 5$
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