Sum of $2008$ consecutive positive integers 
The sum of $2008$ consecutive positive integers is a perfect square. What is the
  minimum value of the largest of these integers?

I understand this means that I need the sum of numbers, where $n=2008$ and I believe that the nearest perfect square might be $1936$ or $2025$. Am I correct? Then, I can equate the sum of $n$ numbers formula to the sum and find the first number in the series, which should be the minimum value, I believe!

The sum is:
$$\sum_{k=0}^{2007} (n+k)=x^2 \Rightarrow 2008n+1004\cdot 2007=x^2.$$
Note that $x=2k$. Then:
$$502n+251\cdot 2007=k^2 \Rightarrow 251\cdot (2n+2007)=k^2.$$
Since $251$ is a prime number, then $k=251m$. Then:
$$2n+2007=251m^2.$$
The smallest $m=3$ and $n=126$. Hence, the minimum of the largest of them is $126+2007=2133$.
