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

First we consider how to represent an odd number. We know that any even number can be represented as 2k for some integer k since, by definition, all even numbers are multiples of 2. We also know that all odd numbers are 1 more than an even number. Thus, any odd number can be written as 2k + 1. Similarly, we know that any multiple of 4 can be written as 4k for some integer k, so any number that is one more than a multiple of 4 can be written as 4k + 1.Now let us consider an arbitrary odd number n = 2m + 1 for some integer m. We aim to square this and rearrange the result into the form 4k + 1. Squaring n gives us n² = (2m + 1)² = (2m + 1)(2m + 1) = 4m² + 4m + 1. This can be written as n² = 4(m² + m) + 1. Since we know that m is an integer, we also know that m² + m is an integer, and therefore n² = 4(m² + m) + 1 is 1 more than a multiple of 4. Since n is arbitrary, this is true for any odd number and therefore we have proven that the square of an odd number is always 1 more than a multiple of 4.