When is \(n^2+2021n\) a perfect square?

Solution

Let \(m,n\in\mathbb{N}\) such that \(n^2+2021n=m^2\). Since \(2021=43\times47\), we have

\[ 4n^2+4\times2021n=4m^2 \]

Adding \(2021^2\) to both sides,

\[ (2n+2021)^2-4m^2=2021^2 \]

Therefore,

\[ (2n-2m+2021)(2n+2m+2021)=2021^2. \]

Case 1: \(2021^2=1\times2021^2\).

Then \(2n-2m+2021=1\) and \(2n+2m+2021=2021^2\). Adding both equations,

\[ 4n+2\times2021=2021^2+1\Rightarrow4n=(2021-1)^2, \]

so \(n=1020100\).

Case 2: \(2021^2=43\times(43\times47)^2\).

Then \(2n-2m+2021=43\) and \(2n+2m+2021=(43\times47)^2\). Solving similarly, we get \(n=22747\).

Hence, \(n^2+2021n\) is a perfect square when

\[ \boxed{n=1020100\ \text{or}\ n=22747}. \]

Here you can download the PDF file (Solution by Thin Sokkean)