 So Brahma Gupta described a procedure where you could solve quadratic indeterminance of the form nx squared plus 1 equals y squared. And Brahma Gupta also notes that the procedure can work if our additive is negative. Now he doesn't give an example, but we can make one. And we'll see why this is an important generalization. So let's try to solve 11x squared plus 1 equals y squared. So as before, we might begin by trying to solve 11x squared plus k equals y squared. And we'll take the easy way out and start with 11 times 1 squared plus something should be a perfect square. And so we think about it. And 5 works 11 times 1 squared plus 5 that's 16 or 4 squared. And so we could try x equals 1, k equals 5, y equals 4. We'll set them down twice and we'll apply Brahma Gupta's procedure. We'll find the Thunderbolt products, that's 1 times 4 plus 4 times 1. We'll find our new additive and we'll find a new last, 11 times the product of the first plus the product of the last. And so remember this gives us a new equation, 11 times 8 squared plus 25 equals 27 squared. And we'd like two things. First, we want the additive to be a perfect square. And it is. But we also need the square root to be a divisor of the first and last. And it isn't. Well, no problem. Remember we can apply Brahma Gupta's procedure over and over again. So if we take our current two rows, we'll find a new first by finding the Thunderbolt products 1 times 27 plus 4 times 8. We'll find a new additive and we'll find a new last. And we'll hear that new additive is not a perfect square. So we'll go on to the next. We'll find a new first by forming our Thunderbolt products. We'll find a new additive and find a new last. And again, this isn't a perfect square. And again, we could keep going, but this doesn't seem to be leading us to a solution. There's a couple problems we're running into. First of all, our additives aren't perfect squares. Our first and last have to be divisible by the square root of the additive. And the additive, since it's a power of 5, will always have a factor of 5. And at first and last, don't seem to have factors of 5. So what can we do? One possibility is we could pick a new value of x, y, and k. And here's where Brahma Gupta's observation is useful. k could be negative. So we might consider 11 times 1 squared minus something gives us a perfect square. And in that case we might find negative 2 works. 11 times 1 squared minus 2 equals 3 squared. Set them down twice. We'll get a new first, a new additive, and a new last. And our additive is a perfect square. And its square root divides both the first and last. And so we can get a solution.