Rational Points and Diophantine Circles

Geometry meets number theory. The unit circle is a continuous curve with uncountably many points on it, but only a special subset have rational coordinates, coordinates expressible as ratios of integers. Listing these rational points turns out to be equivalent to listing Pythagorean triples, the integer solutions of $a^2 + b^2 = c^2$. This connection between a geometric question (rational points) and an arithmetic one (integer triples) is one of the entry doors to algebraic number theory.

What is a rational point?

A point $(x, y)$ on a curve is a rational point if both $x$ and $y$ are rational numbers (elements of $\mathbb{Q}$). Rational points are extremely sparse compared to the full continuum, but on the unit circle, there are infinitely many, and we can list them all.

The Pythagorean connection

Suppose $(p/c, q/c)$ is a rational point on the unit circle $x^2 + y^2 = 1$ with $p, q, c$ integers. Then

$\frac{p^2}{c^2} + \frac{q^2}{c^2} = 1 \quad\Longrightarrow\quad p^2 + q^2 = c^2.$

That is precisely the equation of a Pythagorean triple. So rational points on the unit circle correspond exactly to Pythagorean triples:

• $(3, 4, 5) \Rightarrow (\tfrac{3}{5}, \tfrac{4}{5})$
• $(5, 12, 13) \Rightarrow (\tfrac{5}{13}, \tfrac{12}{13})$
• $(8, 15, 17) \Rightarrow (\tfrac{8}{17}, \tfrac{15}{17})$
• $(7, 24, 25) \Rightarrow (\tfrac{7}{25}, \tfrac{24}{25})$

• Cryptography: Elliptic-curve cryptography looks for rational points on cubic curves; the same techniques that find rational points on a circle (chord-tangent method) extend, and the difficulty of finding all such points on ECs underpins ECDSA.
• Number Theory: Pythagorean triples (3, 4, 5), (5, 12, 13), and so on are in one-to-one correspondence with rational points on the unit circle, the parametrisation $\left(\tfrac{1-t^2}{1+t^2}, \tfrac{2t}{1+t^2}\right)$ generates them all.
• Exact Arithmetic: When a calculation must be exact (e.g., legal land surveys), engineers prefer rational points to floating-point approximations; finding rational points on circles is the geometric side of this preference.

The widget shows the unit circle with a handful of famous Pythagorean triples marked as rational points $(a/c, b/c)$. Drag the slope $t$ to sweep a line from $(-1, 0)$ across the circle; the intersection point traces out every rational point as $t$ ranges through $\mathbb{Q}$. Pick a preset triple from the dropdown to jump the line to the matching $t$-value.

The parametrisation (all rational points at once)

Here is the magic: there is a single formula that generates every rational point on the unit circle as we vary a single rational parameter $t$:

$x = \frac{1 - t^2}{1 + t^2}, \qquad y = \frac{2t}{1 + t^2}.$

You can verify $x^2 + y^2 = 1$ by algebra. The only point missed is $(-1, 0)$ (which corresponds to "$t = \infty$"). For rational $t = p/q$, the coordinates simplify to give a Pythagorean triple. The construction comes from intersecting the line through $(-1, 0)$ of slope $t$ with the circle.

Scaling and translating

To find a rational point on a different circle $(x - h)^2 + (y - k)^2 = r^2$ with rational $h, k, r$: start from a rational point $(p/c, q/c)$ on the unit circle, scale by $r$, and translate by $(h, k)$:

$\Big(h + r \cdot \tfrac{p}{c},\; k + r \cdot \tfrac{q}{c}\Big).$

This stays rational as long as $h, k, r$ are. So the moment you know the rational points on the unit circle, you know them for every rational circle.

Try it

• Plug $t = 1/2$ into the parametrisation: $x = (1 - 1/4)/(1 + 1/4) = 3/5$, $y = 1/(5/4) = 4/5$. The $(3, 4, 5)$ triple.
• Plug $t = 2$: $x = (1 - 4)/(1 + 4) = -3/5$, $y = 4/5$. The same triple, but reflected.
• Plug $t = 3$: $x = (1 - 9)/10 = -4/5$, $y = 6/10 = 3/5$. Same numerators, different positions.

A trap to watch for

It is tempting to think that "most" points on a circle are rational. Wrong by a wide margin: the rational points form a countable set inside the uncountable continuum of all points on the circle. The vast majority of points have irrational coordinates, for example $(\tfrac{1}{\sqrt 2}, \tfrac{1}{\sqrt 2})$ lies on the unit circle but is irrational. Equally important: not every circle has any rational points at all. The circle $x^2 + y^2 = 3$ has no rational points.

What you now know

You can list rational points on the unit circle (and on any rational-radius rational-centre circle) via Pythagorean triples, and you can generate all of them with the rational parametrisation. The next chapter introduces geometric transformations of points, dilations, reflections, and additions, that turn the plane itself into an algebraic object.

Quick check

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References

• Lang, S. (1971). Basic Mathematics. Springer. Chapter 8, §4, rational points and the Pythagorean parametrisation.
• Silverman, J. H., and Tate, J. T. (2015). Rational Points on Elliptic Curves, 2nd ed. Springer. Chapter 1 generalises this entire story.
• Hardy, G. H., and Wright, E. M. (2008). An Introduction to the Theory of Numbers, 6th ed. Oxford. §13.2 derives the Pythagorean triple parametrisation classically.
• Stillwell, J. (2010). Mathematics and Its History, 3rd ed. Springer. Chapter 1 traces Pythagorean triples back to Babylonian tablets.