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
Best Answer
In each four consecutive integers there is exactly one pair of even integers, and exactly one of them is $\;2\pmod 4\;$ and the other one is $\,0\pmod 4\;$ , so the product of these two even (consecutive even) integers is already divisible by $\,8\,\ldots\ldots$