 So to recap what we've done so far, let R be a primitive nth root of unity and designate R to power G to be bracket G. If the sequence whose indices are powers of G has F distinct values, we use the notation F lambda to be those indices multiplied by lambda. And our goal is to make F lambda the solution to an equation. Unfortunately we can't typically do that. What we can do is to make all the possible values of F lambda the solutions to an equation. So we want to make the values of F lambda to be the solution to an equation. So suppose F1, F lambda 1, F lambda 2, and so on are the distinct values. The equation with these solutions will be Z minus F1 times Z minus F lambda 1 and so on. And if we expand, we'll end up with an equation in Z whose coefficients are sums of products of the F lambda i's. And so the question you've got to ask is, what do we know about these sums and products? And Gauss proved an important result. The product F lambda F mu will consist of F summands of the form F nu. So for example, let R be a primitive seventh root of unity. Find 2 1 times 2 2. So earlier we found these values. 2 1 is bracket 1 plus bracket 6. 2 2 is bracket 2 plus 5. 2 3 bracket 3 plus 4. So we can expand 2 1 times 2 2. Well that's bracket 1 plus 6 times bracket 2 plus 5. And we can multiply these out. And remember the useful thing about brackets is if I multiply two bracketed numbers, it's just the sum of the two indices. And we can simplify this. Because we're dealing with seventh roots of unity, then we know that the seventh powers can be dropped. And so this simplifies to... And we'll rearrange this a little bit so that our first two terms are going to be 2 1 and the remaining terms are going to be 2 3. The important thing here is that once we've found the values of any of these numbers, we know their products are going to be expressible in terms of the numbers themselves. So let's work our way through the problem of finding the fifth roots. So let R be a primitive fifth root of unity. And our initial goal is to find a period that does not include all of the roots. So we note the period beginning 1 2 includes 4 and 3 and so this includes all the roots. The period beginning 1 3 includes all the roots. The period beginning 1 4 only includes two of the roots. So we want to start with that period and using Gauss's notation 2 1 is that sum bracket 1 plus bracket 4. And since bracket 2 is not one of the roots we have so far, then we can find 2 2 which will be... And since we're dealing with the fifth roots, this set includes all of those roots. So now let's take the equation with 2 1 and 2 2 as roots. And this equation will be... And we can expand that. So 2 1 plus 2 2, well we know what 2 1 and 2 2 are, so this coefficient will be... This product, well we can figure out what that product is, that's 1 plus 4 times 2 plus 3, which is... Of course this doesn't seem to be useful because if we don't know the primitive roots, how do we know what these sums are? But remember that if r is a primitive nth root of unity, then the sum of the powers plus 1 is equal to 0. And equivalently if we add up all the powers we should get negative 1. And so we can replace this sum 1 plus 2 plus 3 plus 4 with negative 1. So replacing and simplifying gives us a quadratic equation and the solutions will be... Now one solution will correspond to bracket 1 plus bracket 4, while the other corresponds to bracket 2 plus bracket 3. Now remember these go all the way back to the sine and the cosine, and so that means we can approximate what these are going to be. And in fact Gauss recommended this procedure, but it turns out it doesn't actually matter which one is which. So we'll assume that one of them, how about the plus square root of 5, is bracket 1 plus bracket 4. So now we know the sum of bracket 1 plus bracket 4, so how can we disentangle them? Well let's take a look at the equation with bracket 1 and bracket 4 as roots. And so that equation will be... And we can expand. And we know what the sum of bracket 1 plus bracket 4 is. And remember this is going to be bracket 5, and because we're dealing with a fifth root of unity, we know that's going to be 1. And again we have a quadratic equation and we can solve. One of the solutions is going to be bracket 1, a primitive root. The other solution is bracket 4, and you should be able to convince yourself that this will also be a primitive root.