n\(⋮̸3\)\(\Rightarrow\left[{}\begin{matrix}n=3k+1\\n=3k+2\end{matrix}\right.\)( \(k\in N\text{*}\))
+) Xét \(n=3k+1\)ta có :
\(n^2-1=\left(3k+1\right)^2-1\)
\(=9k^2+6k+1-1\)
\(=3k\left(3k+2\right)⋮3\)
+) Xét \(n=3k+2\)ta có :
\(n^2-1=\left(3k+2\right)^2-1\)
\(=9k^2+12k+4-1\)
\(=3\left(3k^2+4k+1\right)⋮3\)
Ta có đpcm