Question 1184724
Maybe two consecutive numbers?

{{{n(n-1)=96}}}



{{{96=4*24=16*6=8*12=2*2*2*2*2*3}}}
?


{{{n^2-n-96=0}}}


{{{n=(1+- sqrt(1^2+4*96))/2}}}

{{{n=(1+- sqrt(385))/2}}}

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Exactly how much "less than the number" do you want?  Maybe you can choose what you need just from looking at factorizations for 96.