 Alright, so let's see how we can use the elliptic curve as a basis for a cryptographic system. And so if we want to use an elliptic curve for a cryptographic system, we have to be able to find points on the curve that are identified with whatever our plain text message is going to be. So we have to find points on the elliptic curve, except the reason that elliptic curves are useful, and the problem with making use of them is that finding points on an elliptic curve is a difficult problem. Now, if we don't mind occasionally having a failure of encryption, Coplitz's original 1987 paper on elliptic curve cryptography did suggest a method for finding points. And so we'll start off by assuming that our messages fall in some interval range. It doesn't really matter what they are, but with Coplitz, he assumed that our messages fell within the range between zero and P over a thousand minus one. Then what we can do is he can find some interval, some value x in the interval, a thousand m to a thousand m plus a thousand. And the idea was that this x is going to correspond to some point on our elliptic curve. And then we can encrypt using whatever method, decrypt using whatever method, and recover m by just ignoring those last three digits. So if m was, say, five, I'd find an x in the interval 5,000 to 6,051,84, for example. And I'll just drop those last three digits of x, and my message is going to be five. And if I want to generalize this, I don't have to use a thousand. I can use some integer k, and I'm going to evaluate, as my decryption, my x component divided by k take the greatest integer that is less than that quotient. For example, I have my elliptic curve y squared equals x cubed plus 2x plus 7 mod 103. And I'm going to identify a suitable value of k to use. And then I'm going to encrypt the message, say m equals 5. Now, since the message is going to be assigned some value of x between k times and k times m plus k, and we're working mod 103, so nothing can be larger than 103. I need to make sure that km plus kb is less than or equal to 103 for all possible values of my message. Now, once we choose k, we limit the possible messages m. And in this particular case, I know that I am sending a message m equals 5. So I have to make sure that whatever k is, 5k plus k is less than or equal to 103. Now, I'll let k equals 14. And again, once I choose my value of k, that limits my possibilities for what my message can be. So here, once I choose k equals 14, working mod 103, my messages have to be someplace between 0 and 6. And if I do have a message between 0 and 6, I'm going to pick x equal to 14m, whatever my message is plus j, where j is someplace between 0 and 14. So by trial and error, I'm going to look for a point that corresponds to a point on the elliptic curve. So my message is 5, I'll look at x equals 70, and we'll check to see if 70 corresponds to a point on the curve. And 103 is congruent to 3 mod 4, so that we know exactly one of the number, or minus the number, has a square root. And y squared congruent to 54 is also congruent to negative 49. So negative 54, 49 has a square root, 54 does not. So we go on to our next point. So our next one, x equals 71. And no obvious square root of either one of those. So I have to try to solve y squared congruent to 32. And it turns out that I do find y congruent to 49 is my solution. So m equals 5, the point 71, 49 is on the elliptic curve. And that's going to be the point that's going to correspond to my message 5. And I'll encrypt this point using some suitable method, possibly elliptic curve elcobal. And at the other end, somebody decrypts it to recover the point 71, 49. They find 71, the x coordinate, divided by 14, my k value. They find the greatest integer less than the quotient, which turned out to be 5. And that's going to be the encrypted message.