multiplication of two consecutive numbers a, Is it true, that if $n \gt 1$ odd number, then we can express  ${n^2-1\over 4}$ as multiply of two consecutive numbers?
b, Is it true, that we can express ${n^4+2n^3+3n^2+2n\over 4}$ in case of n= all natural numbers, as multiply of two consecutive numbers?
What I did already: I know, that the multiply of two consecutive numbers can end to 0,or 2, or 6.
a. ${n^2-1\over 4} = {(3+2m)^2-1\over 4}= {9+2*3*2m+4m^2-1\over 4} = {8+12m+4m^2\over 4}=2+3m+m^2$
m= 0,1,2,3...

a. $\frac{(2m+1)^2-1}{4}=m(m+1)$.
b. $\frac{n^4+2n^3+3n^2+2n}{4}=\left(\frac{n(n+1)}{2}\right)\left(\frac{n(n+1)}{2}+1\right)$
