What is going on algebraically here? The algebra is not hard, but a little dense; follow this closely.

• Algebraically, if \(x^2-2y^2=1\), then the tangent line at any point \((x_0,y_0)\) is given by implicit differentiation to be \(y'=\frac{x_0}{2y_0}\).
• What is the line through \((1,0)\) with that slope? It's \[y=\frac{x_0}{2y_0}(x-1)\; , \] of course.
• So let's check where else this intersects the hyperbola, if at all.
  • We start off with \[x^2-2y^2-1=x^2-2\left(\frac{x_0}{2y_0}(x-1)\right)^2-1=\] \[\left(1-\frac{x_0^2}{2y_0^2}\right)x^2+\left(\frac{x_0^2}{y_0^2}\right)x+\left(-1-\left(\frac{x_0^2}{2y_0^2}\right)\right)=0\, .\]
  • This can be simplified to \[(2y_0^2-x_0^2)x^2+2x_0^2 x+(-2y_0^2-x_0^2)=0\]
  • This, unbelievably, gives us (via the quadratic formula or factoring out \(x-1\)) \[x=\frac{-2x_0^2-4y_0^2}{-2x_0^2+4y_0^2}=\frac{x_0^2+2y_0^2}{x_0^2-2y_0^2}=x_0^2+2y_0^2\]
  • So with a slick substitution of the original point, \[y=\frac{x_0}{2y_0}(x-1)=\frac{x_0}{2y_0}(x_0^2+2y_0^2-(x_0^2-2y_0^2))=2x_0 y_0\, .\]

Now let's try this with actual points!

Awesome!

Brahmagupta knew how to do this, though of course he did it both without our geometric interpretation (which was only made possible by Descartes and Fermat's introduction of coordinate systems) and also without the benefit of symbolically representing \(\sqrt{2}\), which provides this alternate description of what we did.

If you were to do the algebra out here, you'd get exactly the same answer as we did above.

Notice that once again we seem to have created a new number system, though this time with a square root of a positive, not negative number! (And yes, it turns out that finding solutions to things like this is related to \(\mathbb{Z}[\sqrt{2}]\cdots\))

This does precisely correspond to multiplying a group element by 2, e.g. \[[5]+ [5]\equiv 3\text{ mod }(7)\text{ is the same type of thing as }(3,2)+(3,2)=(17,12)\; .\] And it turns out that there is a more general formula that corresponds to taking the line through two points and then moving it so that it goes through the original point \((1,0)\):

If \((x_1,y_1)\) and \((x_2,y_2)\) are both solutions of \(x^2-2y^2=1\), then so is \[(x_1 x_2+2y_1 y_2, x_1 y_2+y_1 x_2)\, \] If you apply this to two points opposite each other like \((3,2)\) and \((3,-2)\), you will get \[(3\cdot 3+2\cdot 2\cdot (-2),3\cdot (-2)+3\cdot 2)=(1,0)\, ,\] which is then the group identity.

This ends up working for any \(n\). Just change all the 2s above to n. Let's see this “by hand” for \(n=3\), where we solve \(x^2-3y^2=1\) with \[2^2-3\cdot 1^2=1\; .\] That is, I use \[x'=x^2+3y^2\text{ and }y'=2xy\]

The general solution, given two points \((x_1,y_1)\) and \((x_2,y_2)\), would be, for \(n>0\) and not a perfect square, \[x'=x_1x_2+ny_1y_2\text{ and }y'=x_1y_2+x_2y_1\; .\]

The same formula works for combining solutions of two different equations like the Pell. \[\text{If }x_0^2-ny_0^2=k\text{ and }x_1^2-ny_1^2=\ell\] \[\text{then }x=x_0x_1+ny_0y_1,\; y=x_0y_1+y_0x_1\text{ solves }x^2-ny^2=k\ell\; .\] This is particularly nice if \(k=\ell=-1\), because it would then give a solution to the Pell equation!

Brahmagupta used this (and more sophisticated things) to solve very hard ones, as did the later English mathematicians who answered some challenges of Fermat:

• Find a nontrivial solution to \(x^2-61y^2=1\).
• Find a nontrivial solution to \(x^2-109y^2=1\).