Proof of the statement “The product of 4 consecutive integers can be expressed in the form 8k for some integer k”

Question

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

Answer 1

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 $\,8\,\ldots\ldots$

Answer 2

Note that if you have four consecutive integers, then one must be divisible by 4 and another must be divisible by 2: they must be equivalent to $0,1,2,\text{ and }3\pmod{4}$.

Answer 3

One way is by proving that the binomial coefficients $$\dbinom n r=\cfrac {n!}{r!(n-r)!}$$ are integers, whoch can be done in various ways including using the binomial recurrence from Pascal's Triangle. Then$$\binom n 4=\frac {n(n-1)(n-2)(n-3)}{4!}$$ which proves that the product of four consecutive integers is divisible by $24$.

Another way is by induction.$$(n+1)n(n-1)(n-2)-n(n-1)(n-2)(n-3)=4n(n-1)(n-2)$$

So you can prove the result using that the product of three consecutive integers is even. Or prove the result for $24$, by showing that the product of three consecutive integers is divisible by $6$ (reducing the number of factors again by taking the difference of successive terms). And remembering to prove the base case - but this can be arranged to have zero as a factor, so divisible by the integer we require.

$$(n+1)n(n-1)-n(n-1)(n-2)=3n(n-1)$$is divisible by 3 and $$(n+1)n-n(n-1)=2n$$ is divisible by 2.

Answer 4

Your formal argument looks fine to me, but you're right: it does seem clumsy and inconvenient. This may be because you are making exactly the same argument in each case: Writing the four numbers, finding one factor of four and another factor of two, and pulling those out to get a factor of $8$.

This suggests a simpler argument. In any four consecutive integers, by the quotient-remainder theorem, there must be one multiple of four. There must also be two multiples of two. One of the multiples of two will also be a multiple of four; the other will not. Therefore, we may factor a four and another two from the product of the four consecutive integers and we see that the product is a multiple of eight.