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