Question

Prove that (n+5)^2-(n+3)^2 is a multiple of 4 for all positive integer values of n

Answer 1

Proved

Explanation:

Given

`(n+5)^2-(n+3)^2`

Required

Show that
`(n+5)^2-(n+3)^2` is a multiple of 4

Expand each bracket

`(n+5)(n+5)-(n+3)(n+3)`

Open brackets

`n^2 + 5x + 5x + 25 - (n^2 + 3x + 3x + 9)`

`n^2 + 10x + 25 - (n^2 + 6x + 9)`

Open bracket

`n^2 + 10x + 25 - n^2 - 6x - 9`

Collect Like Terms

`- n^2 + n^2 - 6n + 10n + 25 - 9`

`- 6n + 10n + 25 - 9`

`4n + 25 - 9`

`4n + 16`

Factorize

`4(n + 4)`

Hence, the multiples of
`(n+5)^2-(n+3)^2` are
`4` and
`(n + 4)`