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

1 Answer
sente
Feb 1, 2016

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.

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