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

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.  

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