\(B=\left(n^2-1\right)\left(n^2-9\right)\)
=(n-1)(n+1)(n-3)(n+3)
\(=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1-3\right)\left(2k+1+3\right)\)
\(=2k\left(2k+2\right)\left(2k-2\right)\left(2k+4\right)\)
\(=16k\left(k+1\right)\left(k-1\right)\left(k+2\right)\)
Vì k;k-1;k+1;k+2 là bốn số liên tiếp
nên k(k+1)(k-1)(k+2) chia hết cho 4!=24
=>B chia hết cho 384