 Alright, well let's see if we can find some perfect numbers now. So we proved previously that if p and q are prime and n is the product of p times q, then n will be a perfect number if and only if p and q are related in this fashion. And so that says I can try and find perfect numbers. Well, maybe. Let's see how this works. So we want to find two primes p and q with the relationship p equals q plus one over q minus one. And so since p can be computed from q, we could just try out different values of q and see what primes p we can produce. So first prime, p equals two. And so if q equals two, then this quotient, q plus one over q minus one, is going to be three over one is going to be three. And remember, p also has to be a prime number, so three works. And so the product two times three also works. So n equals six, which is a perfect number. Well, that worked out pretty well. So let's try our next prime number. So if we could have q equals three, and so if q is equal to three, then q plus one that's four, q minus one two, four over two gives us p equals two, and n is the product three times two. Again, n equals six is a perfect number, but it's not a different one. Well, our next prime q equals five, and q plus one over q minus one, five plus one over five minus one, that's six over four, except that isn't an integer, so it has no hope of being a prime number. And actually we have this problem. We could keep trying different values of q to find additional primes p, but after a while we'd find that nothing really works. So again, let's be suspicious that maybe there aren't any other perfect numbers that are the product of two primes. Well, let's see where we go with that. So let's try some deduction here. I need q plus one over q minus one to be p, so let's try a little bit of algebra here. That means I need q plus one to be the product of two numbers p and q minus one. If in doubt we can multiply things out, so here's something that may be useful to do. q plus one is q minus one plus two, and the reason that I want to do this is that gives me a q minus one, and I already have a q minus one, which gives us some hope of doing a factorization. So let's rearrange our terms and we find that two is p minus one times q minus one. Now I'm not going to do the full algebraic derivation here. I suggest you convince yourself that if I do that I do actually get two as the product p minus one times q minus one, and that tells me that two is the product of two numbers. Well two is also prime, which means that one of these numbers is one, and the other one is going to be two. So either p minus one is one, and q minus one is two, which gives us the solution p equals two, q equals three, or vice versa. p minus one is two, q minus one is one, and again p is three, q is two, and in both cases we get n equals six as our perfect number. Now remember that we started with the idea that n is the product of two primes. All of this work tells us that if n is the product of two primes, it has to be six. And so I can join my first statement, n is the product of two primes, so my last statement, n has to be six, and I end up with the following conclusion. First off, if n equals q plus one over q minus one, the primes have to be two and three, and if I put everything together, that tells me that n is six, and in the case where n is the product of two primes and n is perfect, the only possibility there is n has to be equal to six. And I can reword this somewhat more positively. n equals six is the only product of two primes, that is also a perfect number. Well, what that means is that if we have a perfect number, it will not be the product of two primes, it will be the product of more than two primes. Well, it may help to consider some additional empirical evidence, so we found six as a perfect number, our next perfect number was 28, so here we have 28 as the product of three prime numbers, two, again, two, and seven. And let's assume we didn't have the patience to find 496, or 8,100 in 28 as the next perfect numbers. Just these two examples suggest that maybe my perfect numbers might be found among things that look like two to some power times a prime number. So let's see where that takes us. As before, we'll start with any assumption we want to, and see what conclusions we can draw from it, in the hopes that we'll actually be able to find something interesting and useful, potentially useful, much more important than it be interesting. So here I'm going to assume that p is a prime number, and n, two to the power k times p is perfect. Well, first off, we can go through our proper divisors of n, they're going to be one, two, two squared, and so on, up to the power two to the k, then p, two p, two squared, p, and so on, up to two to the power k minus one times p, and the sum of the proper divisors, well, add all of those things together. And the first thing that's worth noting is this first portion of the sum is the geometric series, and I know how to add together all the terms of the geometric series, it looks like that. The second part is p times another geometric series, and again, I know how to add together the terms of the geometric series, it looks something like that, and after all the dust settles, I end up with a nice similar expression for what the sum is. Now, since n is a perfect number, this sum should be equal to the number itself, and I can do a little bit of algebra. So I'll rearrange this, and I get p equals two to the power k minus one, and joining the starting statement, p is prime, and n equals two to the k, p is perfect, to the ending statement, I now have an important conclusion that I can draw. If I have a perfect number of the form two to the k times p, p itself is a prime number of the form two to the k plus one minus one. Which is a nice conclusion, except in terms of creating perfect numbers, it's set up the wrong way, because again, what I have is I'm starting with the perfect number, and now I know something about it. What I actually want is to start with something else, and then say, aha, here we have a perfect number. Fortunately, our proof is 100% reversible, because again, it's based on algebra, although again, we need to actually go through these steps to verify that we can reverse all of them. And what that gives us is the following information. If p is prime, n equals two to the power k, p is perfect, if and only if, p is of the form two to the k plus one minus one. And somewhat more elegantly, we can let n be k plus one. And so that gives us p equals two to the n minus one, and then what's neat here, this value k here is going to be n minus one. And so we have our final replacement here. n equals two to the power n minus one, two to the n minus one is perfect, if and only if, two to the n minus one is a prime number. And these form what are called the Euclidean perfect numbers. And so I can find them pretty straightforward. I try at different values of n, and I find if n equals two, two to the two minus one, three is prime, so six is perfect. If n equals three, two to the third minus one, seven, prime number, so 28 is perfect. n equals four, not prime, go on to the next. Prime gives me 496, 63, not prime, go on to the next. 127 prime, 81, 28 is a perfect number. Not prime, go on to the next. Not prime, go on to the next. Not prime, go on to the next. Not prime, go on to the next. Not prime, finally we get the next prime, 81, 91. So two to the 13 minus one times 81, 91, 33, 5, 53, 36 is our next perfect number. And so we have our Euclidean perfect numbers. A good proof raises new questions, and in this particular case, it's a lot of work determining whether or not we get a prime number, and it might be useful if we knew when we produced prime numbers. And so the next question that we might want to ask, when is two to the power n minus one going to be a prime number? And we'll take a look at that next.