 So, while Diophontus invented a form of notation, it was primarily for bookkeeping purposes so that he could solve some extraordinary problems. The basic problem solved by Diophontus is the following. Find rational numbers that satisfy a certain relationship. Today, such problems are known as Diophontine equations. Now, the arithmetic actually begins with some non-Diophontine problems. We'd consider them to be standard algebra problems. However, Diophontus' approach is noteworthy. Because he only had one symbol for the unknown, all quantities had to be expressed in terms of it. So, for example, the very first problem of the arithmetic to divide a given number into two, having a given difference. And the problems in Diophontus are generally stated in this way. There's a particular situation and the problem is solved by presenting an example. In this case, Diophontus says given number 100, given difference of 40. Now, the solution to almost every problem in the arithmetic begins with Diophontus saying something to the effect of let the two numbers be followed by what we would consider to be an algebraic expression. Since Diophontus has no other symbol for the unknown, the two parts must be expressed in terms of the single unknown, which will represent as X. And typically the expression that Diophontus is starting with is going to satisfy at least one of the conditions of the problem. In this case, since the difference is 40, then the two numbers must be X and X plus 40. And so Diophontus' first line is let the numbers be X and X plus 40. Now, we have a second condition, which is that our given number is going to be divided into two parts. And since the two numbers together make 100, then we know that X plus X plus 40 is 100. And so the next line in Diophontus' solution is the observation that 2X plus 40 is 100, at which point he solves it more or less as we would solve it, 2X is 40 and X is 30. And here's an important feature, the two numbers that we're looking for are X and X plus 40. We've found X, but we haven't found the two numbers. So we're going to replace X in our expressions and get the required numbers 30 and 70. Book 2 is where the Diophantine problems really begin. And probably the most important problem in book 2 is problem 8, to divide a given number into two squares, given square number 16. It helps to think of the conditions we are requiring. So first, the two numbers add to 16. Second, the first number is square. And the second number is square. We want our expression for the unknowns to satisfy as many of these conditions as we can, then use the remaining condition to set up an equation. And so we might proceed as follows. If the first number is X squared, we automatically satisfy this second condition. Now, we don't have a second variable we can use for the unknown, so we can't make our second number a square. But we could make our second number 16 minus X squared, which will satisfy the first condition because our two numbers will add to 16. And so Diophontas begins, let the numbers be X squared and 16 minus X squared. So again, it's worth noting that by our choice of our variable expressions, we've satisfied two of the three requirements for the problem. We just need to satisfy that third requirement. And so we require 16 minus X squared to be square. And this is the point where solving a Diophonetian equation is an art more than a science. We want 16 minus X squared to be a square. And so Diophontas says, let it be the square of any number of X minus 4, say the square of 2X minus 4. And at this point, we all say, what? Where did this 2X minus 4 come from? And the answer to this question is we get to pick the expression. So we chose an expression that's going to make it possible to solve our problem. So let's think about that. Y 2X minus 4. And our analysis might look something like this. Since we need a square to equal 16 minus X squared, the constant term of what we're squaring must be 4 or negative 4. We need an X term, so we choose 2X arbitrarily. While we could have chosen 2X plus 4, note that if X is positive, this would have given us a number more than 4. But we want two squares that add to 16, so both have to be less than 4. So we'll want to use 2X minus 4. Now, Diophontas doesn't use negative numbers, but we do. And it turns out it doesn't make a difference for us as long as we're willing to handle negative numbers and negative solutions. Still, to remain true to the spirit of Diophontas, we should always make sure that in the end our actual solutions are non-negative rational numbers. So continuing, we want 16 minus X squared to be the square of 2X minus 4. So we can expand and solve. And that gets us our value for X. And again, X is not necessarily what we're trying to solve for. The two numbers are X squared and 16 minus X squared. So our two squares are going to be... Now, it's worth pointing out that if we had chosen different expressions, we'd get different answers. Diophontas only looked for one solution, but it's clear that different choices would give us different answers. For example, if we had chosen that 16 minus X squared was the square of, say, 5X minus 4, we'd get. And so we'd get our solutions. X squared is 4169 and 16 minus X squared is the square of 4013. Again, if we want to remain true to the spirit of Diophontas' work, whatever we get for X, our final answers should be non-negative rational numbers.