Now we can use our geometric intuition to reveal what is happening algebraically here. The algebra is not hard, but a little dense; follow this proof closely.
Proposition15.6.1.
Doubling integer points on the hyperbola \(x^2-2y^2=1\) yields more integer points.
Algebraically, if \(x^2-2y^2=1\text{,}\) then the tangent line at any point \((x_0,y_0)\) other than \((\pm 1,0)\) is given by implicit differentiation to be \(y'=\frac{x_0}{2y_0}\text{.}\) So we start there.
What is the line through \((1,0)\) with that same slope? It’s
To recap, given a point \((x_0,y_0)\) we have achieved a new point \((x_0^2+2y_0^2,2x_0 y_0)\text{.}\)
Now let’s try this with actual points in Sage! I have provided both a numerical and a graphical interact.
Awesome!
Subsection15.6.2Yet more number systems
As mentioned earlier, Brahmagupta knew how to do this. Of course, he did it both without our geometric interpretation (which was only made possible by Descartes and Fermat’s introduction of coordinate systems, though at least Fermat when he examined these was not thinking geometrically) and also without the benefit of symbolically representing \(\sqrt{2}\text{,}\) which provides this alternate description of what we did.
Fact15.6.2.
If \((x_0,y_0)\) is a solution to \(x^2-2y^2=1\text{,}\) then so is \((x_1,y_1)\) where
If you were to do the algebra out here to get a formula for \((x_1,y_1)\) in terms of \((x_0,y_0)\text{,}\) you’d get exactly the same answer as we did above (Exercise 15.7.16).
Moreover, notice that once again we seem to have created a new number system, though this time we have added to the integers the square root of a positive, not negative number! (And yes, it turns out that finding solutions to this equation is related to \(\mathbb{Z}[\sqrt{2}]\cdots\text{.}\) 11 )
Furthermore, the “point doubling” procedure precisely corresponds to multiplying a group element by 2. That is to say:
\begin{equation*}
[5]+ [5]\equiv 3\text{ (mod }7)\text{ is the same type of operation as }(3,2)+(3,2)=(17,12)\text{.}
\end{equation*}
It turns out that there is a more general formula that corresponds to taking the line through two (integer) points and then creating a line with the same slope that goes through the original point \((1,0)\text{:}\)
Example15.6.3.
If both \((x_1,y_1)\) and \((x_2,y_2)\) are solutions of \(x^2-2y^2=1\text{,}\) then so is
In this sense, if we treat \((1,0)\) as an identity element in the sense of group identity, then \((3,-2)\) may be considered the additive inverse of \((3,2)\text{.}\)
This procedure ends up working for any \(n\text{.}\) Just change all the \(2\)s above to \(n\)s. Let’s see this “by hand” for \(n=3\text{,}\) where we solve \(x^2-3y^2=1\) with
This is particularly nice if \(k=\ell=-1\text{,}\) because getting a solution for that would then give a solution to the Pell equation!
Brahmagupta used analogous techniques for his time (and more sophisticated things) to solve very hard ones, as did the later English mathematicians who answered the following challenges of Fermat.
which we can check below. The great mathematician André Weil [E.5.8, II.XIII] finds that Fermat’s comment to his counterparts that the numbers \(61\) and \(109\) were ones selected to not give too much trouble was ‘mischievously’ said; do you agree?
Considering that Brahmagupta says that finding the solution \(x=1151,y=120\) to the equation \(x^2-92y^2=1\) within a year proved the person “was a mathematician”, we can be very thankful for computers!
For more details connecting the topics of this section more directly to abstract algebra, see [E.2.7, Sections 5.3-5.4]; for a more geometric viewpoint, see the same author’s Numbers and Geometry, Chapters 4 and 8.