JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $g cd (m, n) = 1$ and $1^{2} - 2^{2} + 3^{2} - 4^{2} + \dots + (2021)^{2} - (2022)^{2} + (2023)^{2} = 1012 m^{2} n$ then $m^{2} - n^{2}$ is equal to $0$ .

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Explanation

Given $g cd (m, n) = 1$ and

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

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

$\Rightarrow (1 - 2) + (3 - 4) + (5 - 6) + \dots + (2021 - 2022) + (2023)$

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

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

$\Rightarrow - (\frac{2022 ( 2022 + 1 )}{2}) + (2023)^{2} = 1012 m^{2} n$

$\Rightarrow - (\frac{2022 \cdot 2023}{2}) + (2023)^{2} = 1012 m^{2} n$

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

$\Rightarrow m^{2} n = 2023$

$\Rightarrow m^{2} n = 1 7^{2} \times 7 (compare both sides)$

$\Rightarrow m = 17, n = 7$

$\Rightarrow m^{2} - n^{2} = (17)^{2} - (7)^{2} = 289 - 49 = 240$

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