If $\operatorname{gcd}(m, n)=1$ and $1^2-2^2+3^2-4^2+\ldots \ldots .+(2021)^2-$ $(2022)^2+(2023)^2=1012 m^2 n$, then $m^2-n^2$ is equal to [JEE Main 2023, 6 Apr (Shift 2)]

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If \(\operatorname{gcd}(m, n)=1\) and \(1^2-2^2+3^2-4^2+\ldots \ldots .+(2021)^2-\) \((2022)^2+(2023)^2=1012 m^2 n\), th…

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