Sequences And Series

If \(\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1\) and \(1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n\) then \(m^{2}-n^{2}\) is equal to :