2023/003) Prove that there exists 2023 consecutive natural numbers whose sum is a perect square

 Let the 2023 consecutive numbers be from n-1011 to n + 1011

For all to be natual numbers $n>=1012$ 

The sum of them = 2023n

$2023 =17 ^2 * 7$

So if we choose n to be of the form $7m^2$ then the sum becomes a perect square

Not $n >= 1012$ or $7m^2 >=1012$ of $m > 13$

So the 2023 number starting from $7m^2-1011$ where $m > =13$ satisfy the criteria