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Section 15.6 The Algebraic Story

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Subsection 15.6.1 Computing the hyperbola

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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.

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

\begin{equation*} y=\frac{x_0}{2y_0}(x-1)\text{,} \end{equation*}

of course. Let's check where else this intersects the hyperbola, if at all.

Start off with plugging the line into the hyperbola:

\begin{equation*} x^2-2y^2-1=x^2-2\left(\frac{x_0}{2y_0}(x-1)\right)^2-1= \end{equation*}
\begin{equation*} \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\text{.} \end{equation*}

This can be simplified and then solved, unbelievably (via the quadratic formula or factoring out \(x-1\)):

\begin{equation*} (2y_0^2-x_0^2)x^2+2x_0^2 x+(-2y_0^2-x_0^2)=0 \end{equation*}
\begin{equation*} 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 \end{equation*}

Finally, do a slick substitution of the original point:

\begin{equation*} 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\text{.} \end{equation*}

To recap, given a point \((x_0,y_0)\) we have achieved a new point \((x_0^2+2y_0^2,2x_0 y_0)\text{.}\)

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Now let's try this with actual points in Sage! I have provided both a numerical and a graphical interact.

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Awesome!

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Subsection 15.6.2 Yet more number systems

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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 2, which provides this alternate description of what we did.

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If you were to do the algebra out here to get a formula for (x1,y1) in terms of (x0,y0), you'd get exactly the same answer as we did above (Exercise 15.7.16).

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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 Z[2]β‹―. 4 )

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.

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Furthermore, the β€œpoint doubling” procedure precisely corresponds to multiplying a group element by 2. That is to say:

[5]+[5]≑3 (mod 7) is the same type of operation as (3,2)+(3,2)=(17,12).

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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):

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Example 15.6.3.

If both (x1,y1) and (x2,y2) are solutions of x2βˆ’2y2=1, then so is

(x1x2+2y1y2,x1y2+y1x2).

If you apply this to two points opposite each other on the same branch of the hyperbola, such as (3,2) and (3,βˆ’2), you will get

(3β‹…3+2β‹…2β‹…(βˆ’2),3β‹…(βˆ’2)+3β‹…2)=(1,0).

In this sense, if we treat (1,0) as an identity element in the sense of group identity, then (3,βˆ’2) may be considered the additive inverse of (3,2).

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This procedure ends up working for any n. Just change all the 2s above to ns. Let's see this β€œby hand” for n=3, where we solve x2βˆ’3y2=1 with

22βˆ’3β‹…12=1.

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That is, I use

xβ€²=x2+3y2 and yβ€²=2xy
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Subsection 15.6.3 The general solution (any n)

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The general solution, given two points (x1,y1) and (x2,y2), would be, for n>0 and not a perfect square,

xβ€²=x1x2+ny1y2 and yβ€²=x1y2+x2y1.

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Even more generally, the same formula works for combining solutions of two different equations like the Pell.

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This is particularly nice if k=β„“=βˆ’1, because getting a solution for that would then give a solution to the Pell equation!

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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.

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Question 15.6.5.

Find nontrivial solutions to these equations:

  • x2βˆ’61y2=1.

  • x2βˆ’109y2=1.

Solution.

Fermat knew that the smallest solution to the second one is

\begin{equation*} x=158070671986249,y=15140424455100\text{,} \end{equation*}

which we can check below. The great mathematician AndrΓ© 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 β€˜mischievously’ said; do you agree?

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Considering that Brahmagupta says that finding the solution x=1151,y=120 to the equation x2βˆ’92y2=1 within a year proved the person β€œwas a mathematician”, we can be very thankful for computers!