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!
On
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 $126+2007=2133$.
Hint: Sum of $2008$ consecutive integers equal to a perfect square can be written as $$n + (n + 1) + (n + 2) + \cdots + (n + 2007) = 2008n + \sum_{i = 1}^{2007}i = x^2$$ You want to find a value of $x$ such that $$\dfrac{x^2 - \sum_{i = 1}^{2007}i}{2008}$$ is an integer.