Prove the square of an odd number is always 1 more than a multiple of 4

We will start by representing an odd number using algebra. Let n be a integer, i.e. 1,2,3,4,etc. then we can define an odd number as (2n)-1. This is because if a number is of the form 2n it must be even, since it can be divided by 2 and still be a whole number. So, if we take 1 away from the even number, i.e. 2n-1, then it must be odd.
Now we will square our odd number. (2n-1)2 = 4n2-4n+1 =4(n2-n)+1.The first term here 4(n2-n) is clearly a multiple of 4 since we have a 4 outside the brackets. We still have the 1 left over, so we have that the square of an odd number is always 1 more than a multiple of 4.

TD

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