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

First we need to find a general form for an odd number, so that we can prove that when we square it, it will be 1 more than a multiple of 4 regardless of which odd number it is. An even number can always be represented as 2n, with n being any number, as this is always divisible by 2 and therefore even. An odd number is always one more than an even number so we can show this as 2n + 1Now we have a general form for an odd number, we can think about what happens when we square it. (2n + 1)² = (2n + 1)(2n +1) = 4n² + 2n + 2n + 1 = 4n² + 4n +1We can rewrite this by taking out a common factor of 4 from the first two terms, giving us: 4(n² + n) + 14(n² + n) is divisible by 4. Which means the answer is always 1 more than a multiple of 4, as required