The positive divisors of the integer n>1n>1 are d1<d2<<dkd_{1}<d_{2}<\ldots<d_{k}, so that d1=1,dk=nd_{1}=1,d_{k}=n. Let d=d1d2+d2d3++dk1dkd=d_{1}d_{2}+d_{2}d_{3}+\cdots+d_{k-1}d_{k}. Show that d<n2d<n^{2} and find all nn for which dd divides n2n^{2}.