Prove that the difference of any two consecutive square numbers is odd

• 3100 views

It is important we first define what we mean by an odd and even number.
An even number is any integer (whole number) number divisible by 2 so we can express any even number as 2x where x is any integer. When counting, every even number is followed by an odd number; 1,2,3... etc.
We can then express any odd number as 2x+1 as it will just be the next number after 2x i.e. add one.
Now any square number can be expressed as n^2 where n is any integer. The next square number can also be written as (n+1)^2 since it will be the square of the next number after n i.e. n+1.
As such, the difference of any two consecutive square numbers can be written as (n+1)^2 - n^2
Expanding this we get (n^2 + 2n + 1) - n^2
This reduces to 2n+1 since the n^2 values cancel.
Since any odd number can be written in the form 2x+1  where x is any integer as earlier defined, 2n+1 is an odd number for any value of n which completes the proof.

Still stuck? Get one-to-one help from a personally interviewed subject specialist.

95% of our customers rate us

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this.