Question

How do you find two consecutive odd integers such that the difference of their squares is 160?

Answer 1

Set up an equation using `2n+1` as the general form of an odd number to find the integers to be `39` and `41`

Explanation:

Any odd integer may be represented as `2n+1` for some integer `n`. Let's let `n` be such that `2n+1` is the lesser of the two consecutive odd integers. Then we have the equation

`(2n+3)^2-(2n+1)^2 = 160`

`=>4n^2+12n+9-4n^2-4n-1 = 160`

`=>8n = 152`

`=> n = 19`

`:. 2n+1 = 39` and `2n+3 = 41`

Thus the two consecutive odd integers with the desired property are `39` and `41`.