Gọi 2 số lẻ liên tiếp là \(n,n+2\)(n lẻ)
Ta có: \(\left(n+2\right)^2-n^2=\left(n+2-n\right)\left(n+2+n\right)\)
\(=2\left(2n+2\right)=4\left(n+1\right)\)
Do n lẻ \(\Rightarrow n+1\) chẵn
\(\Rightarrow n+1⋮2\)
\(\Rightarrow\left(n+2\right)^2-n^2=4\left(n+1\right)⋮4.2=8\left(đpcm\right)\)

