Prove that the square of an odd number is always 1 more than a multiple of 4
We start off by defining what an odd number is. We take a general letter, say the letter n, to represent any number. If we multiple it by 2 we are sure no matter what number we enter as n the answer will always be an even number. As you can see in the question we want to deal with odd numbers so we add a one to '2n' and we will always get an odd number. As giving in the question we will have to square '2n+1'.
We expand the brackets like so and be careful with the multiplication. The result we get from this is '4n2 + 4n +1' and we look back at the question and remember what it is asking us. To prove that part of this is a multiple of 4 we divide '4n2 +4n' by 4 and pull it out of the bracket we make to get altogether '4(n2 +n) +1'. Now you can see that we have solved the problem as n can be any number being multiplied by 4 then a 1 is being added.
