Starting from Fermat's Last Theorem (12): Revisiting the Tangent Line Method

从费马大定理谈起（十二）：再谈谈切线法 苏剑林 · 2014-10-25 · Mathematics geometry number theory Diophantine equation fermat's last theorem rational point

First, a brief digression about this series and about how this blog is written. In fact, the content of this blog reflects whatever research I happen to be doing around the time I'm writing it. In other words, if I'm writing an article right now, that means I'm currently investigating this particular problem. Whether the subsequent research will yield results, and what kind of results, is something I genuinely don't know yet. So when I write a given article, I'm not sure what the next one will be about. Some related topics I group into the same series. But regardless, these articles may not follow the usual pedagogical or study-oriented logic, and some of the content may deviate considerably from mainstream approaches — please bear with me, and thank you for your continued support!

In the previous post we discussed the tangent-line method for finding rational points on quadratic and cubic curves. This method works very well for finding rational points on curves of degree at most three, but is there an analogous method for curves of higher degree? In other words, is there room to generalize it? Let's revisit, from a purely algebraic point of view, why the tangent-line method works in the first place. The tangent method — or more generally, the secant method — works mainly because if a cubic equation with rational coefficients has two rational roots, then the third root is necessarily rational too. If we only know one rational root, we can make two of the roots coincide with that known root, turning the secant into a tangent. more

Now let's consider quartic curves. Note that for a quartic equation with rational coefficients, we need to have three rational roots in hand beforehand in order to guarantee that the fourth root is rational. Three rational roots correspond to three rational points. We know that two points determine a line, so three points will generally only determine a parabola (or some other type of curve — in any case, this curve needs to have three free parameters in order to "interpolate" these three points. The curve we fit must also satisfy the following: substituting a rational independent variable must always yield a rational dependent variable. For example, with $y=x^2$, substituting a rational $x$ gives a rational $y$; but with something like $y^2=x^2+1$, substituting a rational $x$ need not give a rational $y$, so this type of curve is not acceptable). What if we only know one rational point? Then that's equivalent to making all three points coincide! In that case, the parabolic secant becomes a parabolic tangent, which is equivalent to the second-order Taylor expansion of the original curve at that rational point! In other words, we're still working with tangents, but not just "tangent lines" — we now also consider more refined "tangent curves."

Let's illustrate this technique by finding the rational points of the Diophantine equation $2x^4=y^2+1$. We already know one obvious rational point, $(1,1)$. Expanding in a Taylor series at this point, we obtain

$$\left\{\begin{aligned}&8x^3=2yy',\quad y'=\frac{4x^3}{y}=4\\ &12x^2=\left(y'\right)^2+y y'',\quad y''=\frac{12x^2-(y')^2}{y}=-4\end{aligned}\right.$$

so the second-order Taylor expansion is

$$y=\frac{1}{2}(-4)(x-1)^2+4(x-1)+1=-2x^2+8x-5$$

The intersection points of this curve with the original curve are found from

$$2x^4=(-2x^2+8x-5)^2+1$$

that is, $x^4-16x^3+42x^2-40x+13=0$. By construction, this quartic equation has a triple root at $x=1$, so the last root must be $x=13$, from which we can solve for $y=-239$, meaning $(13,239)$ is another positive integer solution. (As it happens, this particular Diophantine equation has only these two positive integer solutions.)

This example might look like our generalization has succeeded, but it would be more accurate to say it illustrates why the tangent-line method fails for higher-degree curves. First of all, our original curve is of degree four, and to carry out the "interpolation" we need at least a parabola, i.e., a curve of the form $y=ax^2+bx+c$. After substituting this into the original curve, in order to still end up with a quartic equation, the degree of $y$ cannot exceed 2! In other words, if we use the parabolic secant or parabolic tangent method, it only applies to quartic Diophantine equations of a very special form, such as $2x^4=y^2+1$ or $3x^4=y^2+xy+1$. This is a highly restrictive condition. If we consider quintic equations, this trick no longer works at all, because we would need four known rational points, and fitting four rational points requires a cubic curve — but a cubic curve, once squared, becomes degree six, which exceeds the original degree. It seems that the parabolic secant and parabolic tangent methods have already reached their limits with quartic Diophantine equations.

Of course, we hope more techniques with clear geometric meaning for finding rational points will emerge.