Problem - 4247
Let $N$ be the product of four consecutive odd numbers. Show that $N\equiv 1\pmod{8}$.
Let these four consecutive numbers be $(2n-3)$, $(2n-1)$, $(2n+1)$, and $(2n+3)$ where integer $n \ge 2$. Then $$\begin{array}{rl} &(2n-3)(2n-1)(2n+1)(2n+3)\\ =&(4n^2-9)(4n^2-1) \\ =&16n^4 - 40 n^2 +9\\ \equiv& 1\pmod{8}\end{array}$$