When is `n^2+2021n` a perfect square?

When is `n^2+2021n` a perfect square?

Solution 

As we knew, `2021=43\times47`

Let find `m;n\inN` such that: `n^2+2021n=m^2`

                            `4n^2+4\times2021n=4m^2`

Or                         `4n^2+4\times2021n+2021^2=4m^2+2021^2`

                            `(2n+2021)^2=4m^2+2021^2`

                            `(2n+2021)^2-4m^2=2021^2`

            `(2n-2m+2021)(2n+2m+2021)=2021^2`

+Case: 1 `2021^2=1\times2021^2`

            `2n-2m+2021=1` and `2n+2m+2021=2021^2`

Sum of these two equations: `4n+2\times2021=2021^2+1`

                        `4n=(2021-1)^2``\rightarrown=1020100`

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

Then,     `2n-2m+2021=43` and `2n+2m+2021=(43\times47)^2`

Solve as the case 1 we will get `n=22747`

Hence, `n^2+2021n` is a perfect square when `n=1020100` and `n=22747`.

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