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

Feb 1, 2016

Set up an equation using $2 n + 1$ 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 $2 n + 1$ for some integer $n$. Let's let $n$ be such that $2 n + 1$ is the lesser of the two consecutive odd integers. Then we have the equation

${\left(2 n + 3\right)}^{2} - {\left(2 n + 1\right)}^{2} = 160$

$\implies 4 {n}^{2} + 12 n + 9 - 4 {n}^{2} - 4 n - 1 = 160$

$\implies 8 n = 152$

$\implies n = 19$

$\therefore 2 n + 1 = 39$ and $2 n + 3 = 41$

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