Discrete Math: Floor Functions

[rd_wingman]

I'm having trouble with this. Can someone help please.

Prove that: for any odd integer n, we have:

floor((n^2)/2) = [(n-1)/2]*[(n+1)/2]

 

[daon]

If n is odd then n^2 is odd and n^2/2 lies miday between the integers (n^2-1)/2 and (n^2+1)/2.

So floor(n^2/2) = (n^2-1)/2.