I am slowly diving into simple number theory and learning how to craft direct proofs. I needed to proof the statement "The product of 4 consecutive integers can be expressed in the form 8k for some integer k".
My approach was to use the quotient-remainder theorem, as it was surely intended in the textbook. The proof looks like follows.
Proof:
Suppose n is any particular but arbitrarily chose integer. We must show that n(n+1)(n+2)(n+3) is divisible by 8. By the quotient-remainder theorem, n can be written in one of the forms 4q or (4q+1) or (4q+2) or (4q+3) for some integer q. We divide into cases accordingly:
Case 1 (n = 4q for some integer q):
$$\begin{align} n(n+1)(n+2)(n+3) & = 4q(4q+1)(4q+1)(4q+2)(4q+3)\\ & = 8[q(32q^3+48q^2+18q+3)] \end{align}$$
Let $m=q(32q^3+48q^2+18q+3)$. Then m is an integer because sums and products of integers are integers. By substitution, $n(n+1)(n+2)(n+3) = 8m$ where m is an integer. Hence n(n+1)(n+2)(n+3) is divisible by 8.
Case 2 (n = (4q+1) for some integer q): $$\begin{align} n(n+1)(n+2)(n+3) & = (4q+1)(4q+2)(4q+3)(4q+4)(4q+5)\\ & = 8[(4q+1)(2q+1)(4q+3)(q+1)] \end{align}$$
Let $m=q((4q+1)(2q+1)(4q+3)(q+1))$. Then m is an integer because sums and products of integers are integers {...} {See case 1, its basically the same reasoning here...}
Case 3 (n = (4q+2) for some integer q): {...}
Case 4 (n = (4q+3) for some integer q): {...}
Conclusion:
I each of the above cases, n(n+1)(n+2)(n+3) was to be shown to be a multiple of 8. By the quotient-remainder theorem, one of these cases must occur, hence n(n+1)(n+2)(n+3) can be written in the form 8k for some integer k. q.e.d. [End Proof]
But honestly, I consider this proof somehow clumsy and too inconvenient [I am a bloody amateur, maybe I am wrong]. Are there any better/shorter ways to prove the above statement?
Thanks in advance. Nikolai