Similarity between spinless fermion and weakly interacting bosons Why do spinless fermions behave in the same way as weakly interacting bosons.
 A: They don't. A very obvious counterexample is the behaviour of alkali atom gases at temperatures low enough for quantum degeneracy to kick in. 
At very low temperatures, bosonic atoms will interact weakly with each other via collisions. There is pretty clear experimental evidence that weakly interacting bosonic atoms form a Bose-Einstein condensate (BEC) at low temperatures. One can also do similar experiments with fermionic atoms, in which case they definitely do not form a BEC. Typically one uses a magnetic trap that only confines one spin state and not the other one, and therefore the system is pretty much perfectly spin-polarised (spinless fermion = spin-polarised fermion for those confused about spin-statistics). Due to the antisymmetry of the fermionic wave function in configuration-space, $s$-wave scattering is completely suppressed for a spin-polarised system. That means that at low temperatures the fermions are virtually non-interacting. 
So in general, a spinless fermion system does not act in any way like a weakly interacting boson system. Perhaps the OP is referring to the well-known equivalence between non-interacting spinless fermions and hard-core bosons (i.e. bosons with an infinitely strong, short-range interaction) in 1D (the Tonks-Girardeau limit). 
