Math Book Cambodia: When is `n^2+2021n` a perfect square?

Monday, March 22, 2021

Solution

As we knew, $2021 = 43 \times 47$ $2021=43\times47$

Let find $m; n \in N$ $m;n\inN$ such that: $n^{2} + 2021 n = m^{2}$ $n^2+2021n=m^2$

$4 n^{2} + 4 \times 2021 n = 4 m^{2}$ $4n^2+4\times2021n=4m^2$

Or $4 n^{2} + 4 \times 2021 n + 2021^{2} = 4 m^{2} + 2021^{2}$ $4n^2+4\times2021n+2021^2=4m^2+2021^2$

${(2 n + 2021)}^{2} = 4 m^{2} + 2021^{2}$ $(2n+2021)^2=4m^2+2021^2$

${(2 n + 2021)}^{2} - 4 m^{2} = 2021^{2}$ $(2n+2021)^2-4m^2=2021^2$

$(2 n - 2 m + 2021) (2 n + 2 m + 2021) = 2021^{2}$ $(2n-2m+2021)(2n+2m+2021)=2021^2$

+Case: 1 $2021^{2} = 1 \times 2021^{2}$ $2021^2=1\times2021^2$

$2 n - 2 m + 2021 = 1$ $2n-2m+2021=1$ and $2 n + 2 m + 2021 = 2021^{2}$ $2n+2m+2021=2021^2$

Sum of these two equations: $4 n + 2 \times 2021 = 2021^{2} + 1$ $4n+2\times2021=2021^2+1$

$4 n = {(2021 - 1)}^{2}$ $4n=(2021-1)^2$ $\to n = 1020100$ $\rightarrown=1020100$

+Case: 2 $2021^{2} = 43 \times {(43 \times 47)}^{2}$ $2021^2=43\times(43\times47)^2$

Then, $2 n - 2 m + 2021 = 43$ $2n-2m+2021=43$ and $2 n + 2 m + 2021 = {(43 \times 47)}^{2}$ $2n+2m+2021=(43\times47)^2$

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

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

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

Math Book Cambodia