\[2n+1\]

where \[n\]

is any number, so let two odd numbers be \[2n+1\]

and \[2m+1\]

.Then

\[\begin{equation} \begin{aligned} (2n+1)^2-(2m+1)^2 &= = (4n^2+4n+1) \\ &-(4m^1+4m+1) \\ &= 4n^2 +4n-4m^2-4m \\ &= 4(n^2+n-m^2-m)\end{aligned} \end{equation}\]

Proved.