Question

Can you prove that the product of any four consecutive positive integers is divisible by four?.

Answer 1

Let $n$ be the smallest of the 4 integers, so the others are $n+1$, $n+2$, and $n+3$. Their product is

$n(n+1)(n+2)(n+3) = n^4 + 6n^3 + 11n^2 + 6n$

Now consider the statement $P(n)$ that says

$4 \mid n^4 + 6n^3 + 11n^2 + 6n$

(i.e. 4 divides the quartic)

Check the base case $P(1)$.

$n=1 \implies 1^4 + 6\cdot1^3 + 11\cdot1^2 + 6\cdot1 = 24$

and 4 | 24 since 24 = 4•6

Now assume $P(k)$ is true. We want to use it to show $P(k+1)$ is also true. This requires proving

$4 \mid (k+1)^4 + 6(k+1)^3 + 11(k+1)^2 + 6(k+1)$

assuming that

$4 \mid k^4 + 6k^3 + 11k^2 + 6k$

Expand the quartic expression completely.

$(k+1)^4 + 6(k+1)^3 + 11(k+1)^2 + 6(k+1) \\\\ ~~~~~~~~ = k^4 + 10k^3 + 35k^2 + 50k + 24 \\\\ ~~~~~~~~ = (k^4 + 6k^3 + 11k^2 + 6k) + (4k^3 + 24k^2 + 44k + 24)$

The first group of terms is divisible by 4 thanks to the assumption $P(k)$. The second group of terms is clearly divisible by 4, since each coefficient is a multiple of 4. So $P(k)\implies P(k+1)$. QED