Topology of Numbers - I

(In this sequence of posts, I will study results from the new book Topology of Numbers. My aim is not look for any particular result, just learn some ideas about ‘numbers’.)

Chapter 0 : A Preview

Pythagorean triples

We want to find all integers such that $ a^2 + b^2 = c^2 $. If we divide by $ c^2 $ on both sides, we see that the problem is equivalent to finding rational solutions to $x^2 + y^2 = 1$; i.e. finding rational points on the unit circle.

We start with a rational point on the circle $ (0,1) $. Now imagine a line through this point $ y = m x + 1 $. Let’s say, it interests the x-axis at point $ (r,0) $. We must have $ r = \frac{-1}{m} $. We can rewrite the equation for the line as $ y = 1 - \frac{x}{r} $. If we look for intersection of this line with the circle,

\begin{align} x^2 + \left(1 - \frac{x}{r} \right)^2 &= 1 \newline x^2 + 1 - 2 \frac{x}{r} + \frac{x^2}{r^2} &= 1 \newline x^2 \left(1 + \frac{1}{r^2} \right) &= 2\frac{x}{r} \newline \implies x &= \frac{2 r}{1 + r^2},\, y = \frac{r^2 - 1}{r^2 + 1} \newline \end{align}

If $ r $ is a rational, so are $ (x,y) $. Conversely, if $ (x,y) $ is rational, the slope $ -\frac{1}{r} $ is rational, and hence $ r $ is rational.

Every arc of the circle contains infinitely many rational points, as they can be mapped to rational points on the x-axis. We say that the rational points are dense in the circle.

If we set $r = \frac{p}{q} $, we get the $x = \frac{2 p q}{p^2 + q^2}, y = \frac{p^2 - q^2}{p^2 + q^2} $, and hence all Pythagorean triples are:

$(a,b,c) = (p^2 - q^2, 2 p q, p^2 + q^2)$

Pythagorean Triples and Quadratic Forms

We can ask, when is an integer a part of a Pythagorean triple? In general, this is asking for integer solutions to the quadratic form,

$A p^2 + B p q + C q^2 = n$

A study of when an integer can be $ c = p^2 + q^2 $ in a triple, leads to the famous Law of Quadratic Reciprocity.

We can also ask, which triples have two elements difference by 1? In particular, $ a - b = p^2 - 2 pq - q^2 = \pm 1 $, can be simplified by defining, $ (x,y) = (p - q,q) $, and we are led to study, $ x^2 - 2 y^2 = \pm 1 $. In general, we can ask for solutions to Pell’s equations,

$x^2 - D y^2 = \pm 1$

Pythagorean Triples and Complex Numbers

Consider Gaussian integers $a + b i $. We can ‘factorize’,

$(a + b i)(a - b i) = c^2$

We can ask if $ a + b i = (p + q i)^2, a - b i = (p - q i)^2 $. Then,

\begin{align} a &= p^2 + q^2 \newline b &= 2 p q \newline c &= p^2 + q^2 \end{align}

Thus, we recover our formula for Pythagorean triples!

Rational Points on Quadratic Curves

We can use similar techniques (connecting known rational solutions by lines), to find all rational points on quadratic curves $ A x^2 + B x y + C y^2 + D x + E y = F $, provided we can find a single solution.

It is possible for curves to have no rational points. Consider $ x^2 + y^2 = 3 $. If a rational solution existed, then we would have solutions to $a^2 + b^2 = 3 c^2 $. We can assume $a,b$ and $c$ have no common factor. Consider the equation mod 4. LHS can be 1 or 2 mod 4. But RHS is 0 or 3 mod 4. Hence, there are no integer solutions.

This technique to check for solutions modulo an integer n is useful since there are only a finite number of possibilities to check. But the implication is not reversible i.e. it is possible for an equation to have solutions modulo n for every n and still have no integer solutions. E.g. $ 2 x^2 + 7 y^2 = 1 $. This obviously has no integer solutions, but we can show (later) that it has solutions modulo n for every n. A useful fact for this proof will be that the ellipse $2 x^2 + 7 y^2 $ has rational solutions, for example, $ (1/3,1/3) $.

Determining when a quadratic curve contains rational points turns out to be much easier than determining when it has integer points. The general problem reduces to finding rational points on ellipses or hyperbolas of the special form $ A x^2 + B y^2 = C $ where $A,B$ and $C$ are integers that are not divisible by a common square other than 1, and such that no two of $A,B$ and $C$ have a prime factor in common. We will later show a theorem by Legendre that $ A x^2 + B y^2 = C $ contains rational points when the following three relations are satisfied:

$AC$ is congruent mod $B$ to the square of some number
$BC$ is congruent mod $A$ to the square of some number
$-AB$ is congruent mod $C$ to the square of some number

E.g. $x^2 + y^2 = 3 $, we need to check if $-1$ is congruent mod 3 to some square. Since this is not possible, the equation has no rational solutions.

Diophantine Equations

They are algebraic equations for which we look for integer solutions.

For quadratic curves e.g. $x^2 + y^2 = n$ , notice that they have either no rational point or infinite dense set of rational points.

Equations of the form $ y^2 = x^3 + a x^2 + b x +c $ are known as elliptic curves, since they occur in the problem of computing the arc length of an ellipse.

Rational Points on a Sphere

Let’s find rational points on $ x^2 + y^2 + z^2 = 1$. We will use an idea similar to the 2D case.Consider the north pole $ (0,0,1) $. Draw a line from the north pole to the $ x y $ plane intersecting it at $ (u,v,0) $. We can parametrize points on the line as

$(x,y,z) = (0,0,1) + t (u,v,-1)$

Substituting in the equation for a sphere, we get

\begin{align} t^2 u^2 + t^2 v^2 + (1 - t)^2 &= 1 \newline t^2 (u^2 + v^2 + 1) - 2 t &= 0 \newline \implies t &= \frac{2}{u^2 + v^2 + 1} \end{align}

Hence,

$(x,y,z) = \left(\frac{2 u}{u^2 + v^2 + 1},\frac{2 v}{u^2 + v^2 + 1},\frac{u^2 + v^2 - 1}{u^2 + v^2 + 1}\right)$

If $u,v$ are rational, so are the co-ordinates of the point given by the above formula. Moreover,the rational points are dense on the sphere as they are dense on the plane.

Challenge Problems

(These are exercises I found harder than usual.)

Find a right triangle whose sides have integer lengths and whose acute angles are close to 30 and 60 degrees. Hint: Find the irrational value $r $ that corresponds to a right triangle with 30 and 60 degrees, then choose a rational number close to $r $.

Solution: For $r = 2 + \sqrt{3} $. If we approximate $ \sqrt{3} \approx \frac{17}{10} $, then we get the triple $ (a,b,c) = (740, 1269, 1469) $. Note that this is a primitive triple, and the angles are approximately 59.8 and 30.2 degrees.

(Easy to prove but an interesting fact) : For every Pythagorean triple $ (a,b,c) $, $abc$ is divisible by 60.

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