You’re sitting at the end of a long conference table, interviewing for your dream job. You’ve made it this far, but there’s just one more question you have to answer.
“Is it possible for a line that passes through the origin to pass through no other rational points?”
Five pairs of intense eyes watch you, waiting for your response. Do you get the job?
You might think this only happens in story problems, but it happened to me. “Rational points” are points in the plane whose coordinates are all rational numbers. For example, $latex left(frac{12}{5},-frac{2}{3}right)$, $latex left(3,frac{1}{2}right)$ and (11, 4) are rational points, but (4, $latexsqrt{2}$) and (π, -1) aren’t, since $latexsqrt{2}$ and π are irrational. Rational points are important to number theorists and cryptographers, and they even lie at the heart of one of the most famous mathematical theorems of all time. But the question before me had to do with a line that passes through the origin, which means it has at least one rational point, namely (0, 0). Could it avoid passing through another? I didn’t know the answer right away, so I had to think about it.
At first it seems like the answer must be no. Around any point in the coordinate plane there are infinitely many rational points close by. With rational points so densely packed, it seems impossible that a line could avoid them all.
But, as it turns out, it is possible.
The key realization comes from thinking about the slope of the line. You may know slope as “rise over run”: This is just the ratio of the change in the y-coordinate (the “rise”) to the change in the x-coordinate (the “run”) as you move along the line. For example, if you were on a line with slope m and you increased your x-coordinate by 1, you would have to increase the y-coordinate by m to stay on the line. That’s how slope works.
Now imagine starting at (0, 0) on a line with slope m. Going 1 to the right and up m will put you on the line at the point (1, m). Thus, if m is rational, this line must pass through another rational point. In fact, the points (2, 2m), (3, 3m), and so on must all be on the line, showing that if a line through the origin has rational slope, it actually passes through infinitely many rational points.
To crack the interview question, you have to consider lines with irrational slopes. Once you do, the answer comes right away. For example, consider the line through the origin with slope $latex sqrt{2}$, which has the equation y = $latex sqrt{2}$x. If a point (a, b) lies on this line, then b = $latex sqrt{2}a$, and as long as a ≠ 0 we can rearrange this as $latex frac{b}{a}$ = $latex sqrt{2}$. But if (a, b) is a rational point then this is impossible: The left-hand side of this equation can’t be rational while the right-hand side, $latexsqrt{2}$, is irrational. So there can’t be any other rational points on this line. (An interesting extension question, which we explore in the exercises at the end of the column: Are there any lines in the plane that pass through no rational points?)
Quick thinking about lines got me the job, but mathematicians have been studying rational points on curves for a long time, and they’re still learning about the complex structure of these points. To get a sense of this, let’s take a look at how rational points work on circles in the plane. For simplicity we’ll just consider circles centered at the origin with radius r, whose equations are always
x2 + y2 = r2.
Some of these circles pass through no rational points. But, curiously, if a circle contains one rational point, then it contains infinitely many. Let’s see why.
Consider the circle centered at the origin with radius 5. This circle has the equation x2 + y2 = 25, and it contains rational points, like (0, 5) and (3, 4), as well as other points, like (2, $latexsqrt{21}$) and ($latexsqrt{11}$, -$latexsqrt{14}$). But knowing just one rational point on the circle can lead us to infinitely many others, and we can use what we just learned about lines to find them.
Imagine a line passing through the rational point (0, 5) on the circle, and let’s assume that the line has a rational slope, like 2. This line will pass through the circle at a second point, and it turns out this other intersection also has to be a rational point.
Some algebra will show us why. Since the line has slope 2 and y-intercept 5, we can write its equation as y = 2x + 5. The equation of the circle is x2 + y2 = 25, so to find the points of intersection, we just solve the following system of equations:
x2 + y2 = 25
y = 2x + 5.
One approach to solving this system is substitution: Since y = 2x + 5, we can substitute 2x + 5 for y in the circle equation to get
x2 + (2x + 5)2 = 25.
You might be tempted to just solve this equation using your favorite technique, but before we do that, let’s make a few observations.
First, this is a quadratic equation. Quadratic equations can have two solutions, or roots, and we already know one of them: Since (0, 5) is a point of intersection of the line and the circle, x = 0 must be a solution to our equation. Notice that this solution, or root, is rational.
Second, since all the numbers in the equation are rational, when we put our quadratic into “standard form” — ax2 + bx + c = 0 — all the coefficients will be rational. A famous result known as Vieta’s formula says that for a quadratic equation in standard form, the sum of the roots is equal to -$latex frac{b}{a}$. In our case this sum is rational, so if one root is rational, the other must be, too.
This means both points of intersection have rational x-coordinates, so the “run” between them is rational. And since the line passing through them has rational slope, their “rise” must be rational as well. This guarantees that the second point of intersection has a rational y-coordinate, making it a second rational point on the circle.
In our example it probably would have been easier to just solve the system of equations and find the other point of intersection, which turns out to be the rational point (-4, -3). But the argument above generalizes very nicely. Imagine any line y = mx + b with rational slope that passes through our circle at a rational point. The system of equations
x2 + y2 = 25
y = mx + b
results in the quadratic equation
x2 + (mx + b)2 = 25
which has all rational coefficients. By the argument above, if the line passes through the circle again, the other point of intersection must also be rational. So if you know one rational point on a circle, you can find infinitely many others just by taking a line with rational slope, passing it through your rational point, and finding the other point of intersection.
A similar approach can be taken to find rational points on so-called elliptic curves. These are “cubic” curves, with equations that have variables raised to the third power. They’re more complex than lines and circles and are of great interest to number theorists and cryptographers. The study of elliptic curves even played a major role in the solving of Fermat’s Last Theorem — a theorem about finding integer points on certain curves that was proved by Andrew Wiles in the 1990s (about 350 years after Pierre de Fermat famously claimed in the margin of a math book that he had a beautiful proof but the margin was too small to contain it).
There are many different kinds of elliptic curves. Here’s a simple example:
y2 = x3 – 4x + 1.
And here is its graph in the plane:
Though it’s not obvious, there are many rational points on this elliptic curve. For example, (0, 1) and (4, 7) both lie on this curve, since
12 = 03 – 4 × 0 + 1
and
72 = 43 – 4 × 4 + 1.
And as with circles, there’s a clever way to use these known rational points to find more on the curve. The secret once again is using lines.
The equation of the line through (0, 1) and (4, 7) is easy enough to find: The slope is the change in y over the change in x, or m = $latex frac{7-1}{4-0}$ = $latex frac{6}{4}$ = $latex frac{3}{2}$, and since the line passes through (0, 1), its y-intercept is 1. This makes the equation of the line through those two points y = $latex frac{3}{2}$x + 1. Here’s the graph of the line along with the elliptic curve:
Notice that the line intersects the elliptic curve three times: at (0, 1), (4, 7) and a third unknown point. It turns out that the third point must also be rational.
Consider the elliptic curve and the line as a system of equations:
y2 = x3 – 4x + 1
y = $latex frac{3}{2}$x + 1.
We can substitute just like we did with the circle and the line. Here’s what we get:
$latex left(frac{3}{2} x+1right)$2 = x3 – 4x + 1.
This is a cubic equation with rational coefficients. A cubic equation can have up to three real solutions, which makes sense, because we are looking for three points of intersection. And we already know two of those solutions: x = 0, which corresponds to the point (0, 1), and x = 4, which corresponds to the point (4, 7).
So we know that two of the three solutions are rational. What about the third? Well, just as it did with our quadratic equation, Vieta’s formula guarantees that the sum of the roots of this cubic equation must be rational, so if two of the solutions are rational, the third must be rational, too. And that solution will lead us to a new rational point on the elliptic curve. It’s not hard to solve our cubic equation and find the third point of intersection to be $latex left(-frac{7}{4},-frac{13}{8}right)$. As with the circle, this argument generalizes: If a line intersects our elliptic curve at two rational points, then if there’s a third intersection, it must also be rational.
This technique is especially useful in combination with another nice feature of this elliptic curve. Notice that the graph is symmetric about the x-axis: The bottom half of the curve is a reflection of the top. Algebraically this means that if the point (a, b) is on the curve, the point (a, –b) must also be on the curve. Thus, if$latex left(-frac{7}{4},-frac{13}{8}right)$ lies on the curve, so does $latex left(-frac{7}{4}, frac{13}{8}right)$.
This symmetry does more than just give us another rational point on the curve. It gives us a new pair of points to run a line through.
The line through the rational points $latex left(-frac{7}{4}, frac{13}{8}right)$ and (0, 1) will yield yet another rational point on the curve. We can reflect that point, repeat our procedure, and continue to find even more rational points. This procedure has been exploited by mathematicians to create elliptic-curve cryptography, a means of creating secret codes that leverages the structure of these rational points on elliptic curves. It’s easy to use this procedure to find rational points on the curve, but if you’re just given a rational point, it’s difficult to figure out exactly where the procedure began. This sort of hard-to-reverse process is essential to making secure secret codes.
The rich structure of rational points on elliptic curves is an active area of mathematical research. Starting with the point (0, 1) we generated an infinite list of rational points on the curve y2 = x3 – 4x + 1, each of which could lead us to the others. But there are more. The rational point (-1, 2) lies on the curve but isn’t on that list, and we can use it to generate a second infinite set of rational points on y2 = x3 – 4x + 1. It turns out that every rational point on this curve is a combination of those two starting points, which means the “rank” of the curve is 2: Two starting points are all that are needed to generate all the infinitely many rational points on the curve. Mathematicians are actively studying the rank of curves like this, and it is currently unknown whether there is a maximum rank possible for elliptic curves.
From the search for integer solutions to equations 2,000 years ago to Fermat’s Last Theorem to elliptic-curve cryptography today, the study of rational points starts with high school algebra and geometry and leads to advanced mathematics in a beautiful and satisfying way. It may even help you land your dream job.
Exercises
1. Give the equation of a line that passes through no rational points.
2. The points (0, 1) and (4, -7) both lie on the elliptic curve y2 = x3 – 4x + 1, and the equation of the line between them is y = -2x + 1. Yet the system of equations
y = -2x + 1
y2 = x3 – 4x + 1
has only two distinct real solutions. Why doesn’t this conflict with the argument given in the column?
3. Show that the circle with equation x2 + y2 = 3 passes through no rational points.
4. What is the third point of intersection between the line through $latex left(-frac{7}{4}, frac{13}{8}right)$ and (0, 1) and the elliptic curve y2 = x3 – 4x + 1? (Warning: only for algebra and arithmetic lovers!)
Answers
Click for Answer 1:
A simple answer is y = $latex sqrt{2}$. Every point on this line will have a y-coordinate of $latex sqrt{2}$, so no point could be rational. A more interesting answer is x + y = $latex sqrt{2}$. Do you see why this second line contains no rational points?
Click for Answer 2:
The line y = -2x + 1 is tangent to the elliptic curve at (0, 1), so the cubic equation (-2x + 1)2 = x3 – 4x + 1 has a repeated root of x = 0.
It’s easy to solve this equation by simplifying and factoring:
(-2x+1)2 = x3 – 4x + 1
4x2 – 4x + 1 = x3 – 4x + 1
0 = x3 – 4x2
0 = x2 (x – 4).
Since the x2 appears in the factored form of the cubic, the solution x = 0 has a multiplicity of 2. In fact, this is how you begin the procedure for finding rational points on elliptic curves: Find a single rational point, then find the equation of the tangent line and see where it intersects the curve again. Since the point of tangency is a double root, the third root that corresponds to the other point of intersection must be rational.
Click for Answer 3:
Suppose the circle passes through the rational point $latex left(frac{m}{n}, frac{p}{q}right)$ where m, n, p and q are all integers, $latex frac{m}{n}$ and $latex frac{p}{q}$ are in lowest terms (i.e., have no factors in common), and n and q are both positive. Then $latex left(frac{m}{n}right)^{2}$ + $latexleft(frac{p}{q}right)^{2}$ = 3.
First, we’ll show that n = q. We have m2 + n2$latexleft(frac{p}{q}right)^{2}$ = 3n2, and since 3n2 and m2 are both integers, so is n2$latexleft(frac{p}{q}right)^{2}$. Since p and q have no common factors, this means that q must divide n. A similar argument using the equation q2 $latexleft(frac{m}{n}right)^{2}$ + p2 = 3q2 shows that n must divide q. Since n and q are both positive integers that divide each other, it must be the case that n = q.
So $latex left(frac{m}{n}right)^{2}$ + $latexleft(frac{p}{n}right)^{2}$ = 3 and thus m2 + p2 = 3n2. Notice that the right-hand side of the equation is divisible by three, so the left-hand side must also be divisible by three. An interesting fact about perfect squares is that every perfect square is either a multiple of three or one more than a multiple of 3. (This can be shown using modular arithmetic.) This means the only way the sum of two perfect squares could be a multiple of three is if each of the perfect squares is a multiple of three.
So m2 and p2 are both divisible by 3. Since 3 is a prime number, this means that m and p are both divisible by 3, and so the left-hand side of the equation is actually divisible by 9. That means the right-hand side of the equation is divisible by 9, and so n must be divisible by 3. But this violates the assumption that $latex frac{m}{n}$ is in lowest terms. Thus there can be no rational point on x2 + y2 = 3.
An interesting, and challenging, extension question is: For what values of k will x2 + y2 = k pass through no rational points?
Click for Answer 4:
$latex left(frac{92}{49}, frac{113}{343}right)$.
The line has the equation y = $latex -frac{5}{14}x$ + 1. To find the x-coordinate, solve the equation $latexleft(-frac{5}{14} x+1right)$2 = x3 – 4x + 1. To be honest, I just had Wolfram Alpha do it.
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Source: https://www.quantamagazine.org/how-simple-math-reveals-rational-points-on-curves-20210722/
