Let $a=(n+k)(n-k+1)<b=(n+k+1)(n-k+1)<c=(n+k)(n-k+2)$ (k>0 integer). Then $a|bc$, if $k(k-1)<n\le k(k+1)$ we have $a>n^{2}$ and $c=n^{2}+2n-k^{2}+2k<n^{2}+n+3\sqrt n$, because $k^{2}-2k>n-3\sqrt n$.