 Let's take a look at a few more problems of Diathontus. The arithmetic books to problem 21 define two numbers So the square of either minus the other is square So let's think about this. We want to find two numbers where one of them squared minus the other is a square So these are really two conditions this relationship has to hold for both of them So if we make our first number x plus one we have x plus one squared minus something should be a square So our goal here is to try to pick our second number to be something where this is automatically true So we know that x plus one squared is x squared plus 2x plus one and if we subtract 2x plus one We'll get a square and what this means is that if we make our two numbers x plus one our first number and 2x plus one our second number will satisfy one of these requirements Automatically the square of one minus the other is a square And so Diathontus begins his solution by saying let the two numbers be x plus one and two x plus one satisfying one condition Now there's a second condition. We've required 2x plus one squared Minus x plus one to be a square the second number squared minus the first has to be a square So the thing we might notice here is that when we expand this out the constant term drops out And so we can say that this difference 2x plus one squared minus x plus one Well, let that be the square nine x squared And so we see that our second and remaining condition has been turned into an equation and we can solve Which gives us a value of x? And again the important thing here is that x is not actually our solution the numbers We're looking for our x plus one and 2x plus one so our two numbers are going to be Now for a variety of reasons one of the most important problems in Diathontus occurs in book 5 problem 9 Where Diathontus considered the problem of dividing 13 into 2 squares both of which were greater than 6 To do so he has to solve the problem of finding a square very nearly equal to 6 and a half And the procedure he follows has been termed Adequating a rough translation of this term might be as nearly equal as possible Now while the problem is to find a square very nearly equal to six and a half We can actually make this problem easier to solve by making it more restrictive And so Diathontus takes a look at the problem to add a square to six and a half to make it a square Now while we could add x squared in recognition of our goal of making our result as close to six and a half as possible We'll let our square be one over x squared And so we want six and a half plus one over x squared to be a square Now this does require us to deal with rational expressions So Diathontus goes to a sequence of manipulations that we can represent as follows If six and a half plus one over x squared then four times this amount 26 plus four over x squared will also be a square or If we let one over y be two over x Then we want 26 plus one over y squared to be a square but if 26 plus one over y squared is a square then so is 26 y squared plus one and So at the end of the day, we want 26 y squared plus one to be the square of something And at this point you might bring up an important observation Diathontus only has one variable. What are we doing introducing y and in fact since Diathontus only has one variable. He begins with the assumption let 26 x squared be a square And so we have to figure out what we want it to be the square of So let's think about that since we want 26 x squared plus one to be the square of something Then it must be the square of something with a constant of one or negative one Also, we might notice that since five squared equals 25. We'll let the variable part be 5x There's no real requirement that this be the case But it turns out that if we do let it be close to the square root. We get much better results So we'll assume that 26 x squared plus one is the square of 5x plus one And we can solve and if x is equal to 10 then we know that 26 times 10 squared plus one equals 5 times 10 plus one squared Now we actually wanted to add something to six and a half, but that's okay. We can work our way back to six and a half And so six and a half plus the square one four hundredth gives us a square five halves plus one twentieth and So one four hundredth is the required square to add to six and a half