2024/042) For each positive integer n we consider the sequence of 2004 integers $\lfloor n+\sqrt{n}\rfloor ,\lfloor n+1+\sqrt{n+1}\rfloor ,\lfloor n+2+\sqrt{n+2}\rfloor,$ $\cdots,\lfloor n+2003+\sqrt{n+2003}\rfloor$. How do I find the smallest integer n, such that the 2004 numbers in the sequence are 2004 consecutive integers?

We know for any integer x

$\lfloor x+y \rfloor = x + \lfloor y \rfloor$

So we have

$\lfloor  n + k + \sqrt{n+k}  \rfloor = n + k  + \lfloor \sqrt{n+k} \rfloor$

From the given 2024 numbers by putting k = 0 t0 2023 we get

 $\lfloor  n +  \sqrt{n}  \rfloor = n +  \lfloor \sqrt{n} \rfloor$

 $\lfloor  n + 1 + \sqrt{n+1}  \rfloor = n + 1  + \lfloor \sqrt{n+1} \rfloor$

$\lfloor  n + 2 + \sqrt{n+2}  \rfloor = n + 2  + \lfloor \sqrt{n+2} \rfloor$

$\lfloor  n + 2023 + \sqrt{n+2023}  \rfloor = n + 2023  + \lfloor \sqrt{n+2023} \rfloor$

They are consecutive if we have

 $ \lfloor \sqrt{n} \rfloor = \lfloor sqrt{n+1} \rfloor \cdots \lfloor \sqrt{n+ 2023} \rfloor$

as these are in increasing order we have

  $ \lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n+ 2023} \rfloor$

The 1st term should be as low as possible so n is a perfect square $x^2$

So we have $ x = \lfloor \sqrt{x^2+ 2023} \rfloor$

 As we have the smallest $k=(x+1)^2$ such that $ x <  \sqrt{k}$

so we have $x^2 + 2023 \lt x^2 + 2x + 1$ or $2024 \lt 2x $ so x = 1013 and

smallest $n = 1013^2= = 1026169$