 If r is a primitive k-th root of unity, Gauss defined bracket k to be r to power k. And for any g, the distinct values in this sequence form what Gauss called a period. And these periods are important because the periods form a natural partition of the roots. One useful simplification. If we think of the indices as formed by repeated multiplication by g, we can simplify as we go along. For example, let r be a primitive seventh root of unity. Find the distinct terms of the sequence beginning bracket one, three. And so we have a sequence where we multiply each index by three. And so our sequence begins bracket one, bracket three times one, or three. Bracket three times three, that's nine. And remember that's really r to the ninth, where r is a primitive seventh root of unity. And because it's a primitive seventh root, we can factor out an r to the seventh, and simplify, or bracket two. Our next term will be bracket three times two, or six. Bracket three times six, that's eighteen, which simplifies to bracket four. The next term is bracket three times four, that's bracket twelve, which simplifies. The next term is bracket three times five, which gives us bracket one, and takes us back to our starting point. And what this means is that since we're going to form the next one by multiplying by three, that'll be bracket three, and our sequence is just going to repeat after that. So suppose bracket one, g, generates a period with f terms. Gauss used the notation f one is the sum of those terms, and f lambda will be the sum where all of our indices are multiplied by lambda. For example, suppose r is a primitive seventh root of unity. Let's find the period beginning bracket one, two. Then use our f lambda notation to express the sum of the roots. Also, let's find f two, f three, and f four. So each index will be multiplied by two to get the next index. So our first two indices are bracket one, bracket two. After that, we get bracket two times two, four. Bracket two times four, that's eight, which simplifies as, which takes us back to our starting point, and so these are the only distinct elements. So the period has three elements, and so we can write this as three one is the sum, bracket one, plus bracket two, plus bracket four. So remember the effect of changing this second value is to multiply the indices by that number. So three two, we're going to take everything and multiply the index by two. So that's going to be bracket two times one, bracket two times two, bracket two times four, or two plus four plus eight. But because we're dealing with seventh roots of unity, this bracket eight can be reduced. And similarly, three three, we're going to multiply every index by three. And simplify. And finally for three four, we'll multiply every index by four. Now if you look at these, you'll notice that some of these are exactly the same, although the order of the sum might be different. But if the terms aren't the same, they are completely different. And so f lambda either has the same terms as f mu or entirely different terms. So let's see if we can find those values. So let R be a primitive seventh root of unity. Let's find all the distinct values of two lambda. Now since we want to find two lambda, we want to find g so that the sequence beginning one g has two distinct values. Now we already found the sequence bracket one two four, which has more than two distinct values. So whatever g is, it's not equal to two. We might try g equal to three and our sequence begins one three three times three. Well that's nine, which reduces down to two. And we already have too many values. If we try g equals four, the first two terms in our sequence are one four then four times four. That's sixteen, which reduces down to two. And again, we already have three values, which is too many. We try five and fail. We try six and we see that six does work because the next term in our sequence is going to be bracket thirty six, which reduces down to bracket one. And so two one, the period with two terms beginning with bracket one, well that's going to be bracket one plus six. Now since f lambda and f mu either have all terms in common or none and f one includes the terms bracket one and six, then we can pick any lambda not already in a set. So we might find two two, that's going to be multiplying each index by two, which gives us. Now we don't have three, so we can find two three, which will be. And now we have all powers of our primitive seventh root and so any other value of lambda would give us a sum we've already found. So why bother? Remember the periods form a natural way to partition the roots and if we can make the values of f lambda the roots of an equation, we can solve the equation and find a certain sums of the roots without knowing the roots. All we have to do is, well, there's a bit more work. We'll take a look at that next.