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 :

Given $\operatorname{gcd}(m, n)=1$ and <br/><br/>$$ \begin{aligned} & \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\ & =1012 m^2 n \\\\ & \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\ & =1012 m^2 n \\\\ & \Rightarrow(1-2)(1+2)+(3-4)(3+4)+\ldots .(2021-2022) \\ & (2021+2022)+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(-1)(1+2)+(-1)(3+4)+\ldots .+ \\ & \quad(-1)(2021+2022)+\left(2023^2\right)=(1012) m^2 n \end{aligned} $$ <br/><br/>$\begin{aligned} & \Rightarrow(-1)[1+2+3+4+\ldots+2022]+(2023)^2 =1012 m^2 n \\\\ & \Rightarrow(-1)\left[\frac{(2022) \cdot(2022+1)}{2}\right]+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(-1)\left[\frac{(2022)(2023)}{2}\right]+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(2023)[2023-1011]=(1012) m^2 n \\\\ & \Rightarrow m^2 n=2023 \\\\\end{aligned}$ <br/><br/>$\begin{array}{lc}\Rightarrow m^2 n=289 \times 7 \\\\ \Rightarrow m^2=289 \text { and } n=7 \\\\ \Rightarrow m=17 \text { and } n=7(\text { such that } \operatorname{gcd}(m, n)=1) \\\\ \therefore m^2-n^2=289-49=240\end{array}$