If gcd(m,n)=1 and 1^2-2^2+3^2-4^2+ ..+(2021)^2-(2022)^2+(2023)^2=1012m^2n then m^2-n^2 is …

Question

If $gcd(m,n)=1$ 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 :

Accepted Answer

Given $gcd(m,n)=1$ and

$\begin{gathered} \Rightarrow {1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+\ldots .+(2021{)}^{2}-(2022{)}^{2}+(2023{)}^{2} \ =1012{m}^{2}n \ \Rightarrow {1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+\ldots .+(2021{)}^{2}-(2022{)}^{2}+(2023{)}^{2} \ =1012{m}^{2}n \ \Rightarrow (1-2)(1+2)+(3-4)(3+4)+\ldots .(2021-2022) \ (2021+2022)+(2023{)}^{2}=(1012){m}^{2}n \ \Rightarrow (-1)(1+2)+(-1)(3+4)+\ldots .+ \ \, (-1)(2021+2022)+({2023}^{2})=(1012){m}^{2}n \end{gathered}$

$\begin{gathered} \Rightarrow (-1)[1+2+3+4+\ldots +2022]+(2023{)}^{2}=1012{m}^{2}n \ \Rightarrow (-1)[\frac{(2022)\cdot (2022+1)}{2}]+(2023{)}^{2}=(1012){m}^{2}n \ \Rightarrow (-1)[\frac{(2022)(2023)}{2}]+(2023{)}^{2}=(1012){m}^{2}n \ \Rightarrow (2023)[2023-1011]=(1012){m}^{2}n \ \Rightarrow {m}^{2}n=2023 \end{gathered}$

$\begin{gathered} \Rightarrow {m}^{2}n=289	imes 7 \ \Rightarrow {m}^{2}=289n=7 \ \Rightarrow m=17n=7(gcd(m,n)=1) \ 	herefore {m}^{2}-{n}^{2}=289-49=240 \end{gathered}$

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