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

In order to prove this we can write a general expression of an odd number in terms of n, e.g - 2n+1Square this 'odd number': (2n+1)^2, therefore you can write it as (2n+1)(2n+1), then expand (multiply out) the brackets to get: 4n^2 + 4n + 1We can then factorise this to get: 4(n^2 + 1) + 1 which is 'one more than a multiple of 4' as 4(n^2 + 1) will always be a multiple of 4 regardless of what n is.