If gcd (m, n) = 1 and 1 2 - 2 2 + 3 2 - 4 2 + . . . + ( 2021 ) 2 - ( 2022 ) 2 + ( 2023 ) 2 = 1012 m 2 n , then m 2 - n 2 is equal to [2023]

Question

If gcd (m, n) = 1 and $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . + {(2021)}^{2} - {(2022)}^{2} + {(2023)}^{2} = 1012$ $m^{2} n$ , then $m^{2} - n^{2}$ is equal to [2023]

Smart Achievers · Accepted Answer

(4)

$We have, (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + \dots + ({(2021)}^{2} - {(2022)}^{2}) + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow \underset{It is an A.P.}{\underset{⏟}{{- 3 - 7 - 11 - \dots - 4043}}} + {(2023)}^{2} = 1012 m^{2} n$

$Now, - 4043 = - 3 + (n - 1) (- 4) \Rightarrow n = 1011$

$So, - (3 + 7 + \dots + 4043) + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow - \frac{1011}{2} [3 + 4043] + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow - 2023 \times 1011 + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow 2023 (2023 - 1011) = 1012 m^{2} n$

$\Rightarrow 2023 \times 1012 = 1012 m^{2} n \Rightarrow m^{2} n = 2023 = 7 \times 17^{2}$

$\Rightarrow m = 17$ and $n = 7$

$∴$ $m^{2} - n^{2} = 17^{2} - 7^{2} = 289 - 49 = 240$

Solution

Q.
If gcd (m, n) = 1 and $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . + {(2021)}^{2} - {(2022)}^{2} + {(2023)}^{2} = 1012$ $m^{2} n$ , then $m^{2} - n^{2}$ is equal to [2023]

1 200

2 180

3 220

4 240

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