Pythagorean Triples, Complex Numbers, Abelian ... - BGU Mathamyekut/lectures/pythagoras/notes.pdf · (4.3) a = u Y1 i=1 qni i; with u 2T, n i 2Z, and all but ˙nitely many n i are

Pythagorean Triples, Complex Numbers, AbelianGroups and Prime Numbers

Amnon Yekutieli

Department of MathematicsBen Gurion University

email: [email protected]

Notes available athttp://www.math.bgu.ac.il/~amyekut/lectures

written 7 June 2015

Amnon Yekutieli (BGU) Pythagorean Triples 1 / 28

mailto:[email protected]

http://www.math.bgu.ac.il/~amyekut/lectures

1. Pythagorean Triples


A Pythagorean triple is a triple (a, b, c) of positive integers, satisfying

(1.1) a2 + b2 = c2.

The reason for the name is, of course, because these are the sides of a rightangled triangle:

a

bc



We say that the triples (a, b, c) and (a′, b′, c′) are equivalent if thecorresponding triangles are similar.

This means that there is a positive number r , such that

(a′, b′, c′) = (ra, rb, rc)

or(a′, b′, c′) = (rb, ra, rc).

Clearly r is rational.

We say that the triple (a, b, c) is reduced if the greatest common divisor ofthese numbers is 1. The triple is called ordered if a ≤ b.

It is easy to see that any triple (a, b, c) is equivalent to exactly one reducedordered triple (a′, b′, c′).

Exercise 1.2. Let (a, b, c) be a reduced ordered triple. Then c is odd, anda < b.



Here is an interesting question:

Question 1.3. Are there in�nitely many reduced ordered Pythagoreantriples?

The answer is yes. This was already known to the ancient Greeks. There is aformula attributed to Euclid for presenting all Pythagorean triples, and itproves that there in�nitely many reduced ordered triples.

This formula is somewhat clumsy, and I will not display it. It can be foundon the web, e.g. here:http://en.wikipedia.org/wiki/Pythagorean_triplehttp://mathworld.wolfram.com/PythagoreanTriple.html

Later we will see a nice geometric argument, essentially based on the Euclidformula, to show there in�nitely many reduced ordered triples.


http://en.wikipedia.org/wiki/Pythagorean_triple

http://mathworld.wolfram.com/PythagoreanTriple.html


For a given integer c > 1, let us denote by PT(c) the set of all reduced orderedPythagorean triples with hypotenuse c.

A more interesting question is this:

Question 1.4. What is the size of the set PT(c) ?

The answer to this was found in the 19th century. We will see it at the end ofthe lecture.

An even more interesting question is this:

Question 1.5. Given c, is there an e�ective way to �nd the reducedordered Pythagorean triples with hypotenuse c ?

An e�ective method (possibly new!) will be presented at the end of the talk.


2. Complex Numbers on the Unit Circle


It was observed a long time ago that Pythagorean triples can be encoded ascomplex numbers on the unit circle.

Starting from a reduced ordered Pythagorean triple (a, b, c), we pass to thecomplex number

z := a + b · i,

that has absolute value c.

Consider the complex number

(2.1) ζ = s + r · i :=z|z|

=ac

+bc· i.



The number ζ has rational coordinates, it is on the unit circle, in the secondoctant, and is di�erent from i.

c

b

a

z

We can recover the number z, and thus the reduced ordered Pythagoreantriple (a, b, c), by clearing the denominators from the pair of rationalnumbers (r, s) = ( ab ,

ac ).



Actually, there are 8 di�erent numbers on the unit circle that encode thesame Pythagorean triple:

(2.2) ± ζ,±i · ζ,±ζ̄,±i · ζ̄.

Figure 2.3.Amnon Yekutieli (BGU) Pythagorean Triples 8 / 28


Given a complex number ζ with rational coordinates on the unit circle, otherthan the four special points ±1,±i, let us denote by pt(ζ) the unique reducedordered Pythagorean triple (a, b, c) that ζ encodes.

In this fashion we obtain a function pt from the set of complex numbers onthe unit circle with rational coordinates (not including the four specialpoints), to the set of reduced ordered Pythagorean triples.

This function is surjective, and it is 8 to 1.

Therefore, to show that there are in�nitely many reduced orderedPythagorean triples, it su�ces to prove:

Proposition 2.4. There are in�nitely many complex numbers on the unitcircle with rational coordinates.



Here is a geometric proof of the proposition.

Let us denote the unit circle by S1.

The stereographic projection with focus at i is the bijective function

f : S1 − {i} → R,

that sends the complex number ζ to the unique real number f (ζ) that lies onthe straight line connecting i and ζ .



Exercise 2.5. Show that ζ has rational coordinates i� the number f (ζ) isrational.

(Hint: use similar triangles.)

Since there are in�nitely many rational numbers, we are done.


3. The Circle as an Abelian Group


Previously we used the notation S1 for the unit circle.

I will now switch to another notation, that comes from algebraic geometry,and is better suited for our purposes.

From now on we shall write

G(R) := S1 = {ζ ∈ C | |ζ| = 1}.

The set G(R) is a group under complex multiplication, because

|ζ1 · ζ2| = |ζ1| · |ζ2| and |ζ−1| = |ζ|−1.



Let G(Q) be the subset of G(R) consisting of points with rationalcoordinates; namely

(3.1) G(Q) = {ζ = s + r · i | s, r ∈ Q, s2 + r2 = 1}.

Exercise 3.2. Prove that G(Q) is a subgroup of G(R).

Recall that to answer Question 1, namely to show there are in�nitely manyreduced ordered Pythagorean triples, it su�ces to prove that the abeliangroup G(Q) is in�nite.



We �rst locate all the elements of �nite order in the group G(Q). These are theroots of 1, namely the elements ζ satisfying ζn = 1 for some positive integern.

Algebraic number theory tells us that there are just four of them:

(3.3) 1, i,−1,−i.

Thus, if we take any element ζ ∈ G(Q) other than those four numbers, thecyclic subgroup that it generates

{ζn | n ∈ Z} ⊂ G(Q)

will be in�nite!



Let us consider the familiar reduced ordered Pythagorean triple (3, 4, 5).

The corresponding number in G(Q) is

ζ := 35 + 4

5 · i,

and it is not one of the four special numbers in (3.3). So this element hasin�nite order in the group G(Q).

Here are the �rst positive powers of ζ , and the corresponding triples.

n ζn pt(ζn) = (an, bn, cn)

1 35 + 4

5 i (3, 4, 5)

2 − 725 + 24

25 i (7, 24, 25)

3 − 117125 + 44

125 i (44, 117, 125)

4 − 527625 −

336625 i (336, 527, 625)



Exercise 3.4. Find a reduced Pythagorean triple with hypotenuse c = 3125.

(Later we will see that there is only one!)

Remark: the algebraic number theory used above, and all that is needed tocomplete the proofs in this lecture, can be found in the book

“Algebra”, by M. Artin, Prentice-Hall.


4. The Ring of Gauss Integers


Consider the ring of Gauss integers

A := Z[i] = {m + n · i | m, n ∈ Z}.

Let us denote its �eld of fractions by

K := Q[i] = {s + r · i | s, r ∈ Q}.

The reason we want to look at K is this: equation (3.1) shows that

(4.1) G(Q) = {ζ ∈ K | |ζ| = 1}.

It is known that the ring A is a unique factorization domain.

There are countably many primes in A. Let us enumerate them as

q1, q2, q3, . . . .



The group A× of invertible elements of A turns out to be the group of rootsof unity

(4.2) T := {±1,±i}.

Unique factorization tells us that any element a ∈ K× can be writtenuniquely a product

(4.3) a = u ·∞∏i=1

qnii ,

with u ∈ T , ni ∈ Z, and all but �nitely many ni are 0 (so the product isactually �nite).

We see that as an abelian group,

(4.4) K× = T × F ,

where F is the free abelian group with basis {qi}i=1,2,....Amnon Yekutieli (BGU) Pythagorean Triples 18 / 28


It is known that the primes of the ring A = Z[i] are of three kinds.

The �rst kind is the primeq := 1 + i.

It is the only prime divisor of 2 in A, with multiplicity 2 :

q2 = (1 + i)2 = 1 + i2 + 2 · i = i · 2.

Next let p be a prime in Z satisfying

p ≡ 3 mod 4.

For example p = 3 or p = 7.

Then p is also prime in A. This is the second kind of primes.



Finally, let p be a prime of Z satisfying

(4.5) p ≡ 1 mod 4.

For example p = 5 or p = 13.

Then there are two primes q and q̄ in A, conjugate to each other but notequivalent (i.e. q̄ /∈ T · q), such that

(4.6) p = u · q · q̄

for some u ∈ T .

The numbers q, q̄ are the third kind of primes of A.

These will be the interesting primes for us.



It is not hard to �nd the decomposition (4.6). The number p is a sum of twosquares in Z :

p = m2 + n2.

We then takeq := m + n · i and q̄ := m− n · i.

Consider the number

(4.7) ζ := q/q̄ ∈ K .

It has absolute value 1, and hence, by (4.1), it belongs to the group G(Q).


5. The Group Structure of the Rational Circle


There are countably many primes of Z that are 1 mod 4.

Let us enumerate them in ascending order:

p′1 := 5 , p′2 := 13 , p′3 := 17 , . . . .

Each such prime p′j of Z has a prime decomposition in the ring A = Z[i] :

p′j = uj · q′j · q′j.

We use it to de�ne the element

(5.1) ζ ′j := q′j / q′j ∈ G(Q).

Thus we get a sequence of elements {ζ ′j}j=1,2,... in the group G(Q).Amnon Yekutieli (BGU) Pythagorean Triples 22 / 28


Here is our main result.

Theorem 5.2.

Any element ζ ∈ G(Q) is uniquely a product

ζ = u ·∞∏j=1

ζ ′jnj ,

with u ∈ T, nj ∈ Z, and all but �nitely many of the nj are 0.

In other words, the abelian group G(Q) is a product

G(Q) = T × F ′,

where F ′ is the free abelian group with basis {ζ ′j}j=1,2,....



Sketch of Proof.

Consider the prime decomposition of ζ as an element of K×, as in (4.3).

For each prime qi of A we calculate its absolute value |qi| ∈ R.

The equality

1 =

∞∏i=1|qi|ni

implies that the primes qi that do not come in pairs must have multiplicityni = 0.

The primes that do come in pairs, namely the primes of the third kind, musthave opposite multiplicities. Thus they appears as powers of thecorresponding number ζ . �


6. Back to Pythagorean Triples


Recall that for a number ζ ∈ G(Q)− T , we write pt(ζ) for thecorresponding reduced ordered Pythagorean triple.

Here is our explicit presentation of all reduced ordered Pythagorean triples.



Theorem 6.1. Let c be an integer greater than 1, with prime decomposition

c = pn11 · · · p

nkk

in Z. Here p1 < · · · < pk are positive primes; n1, . . . , nk are positive integers;and k is a positive integer.

1. If pj ≡ 1 mod 4 for every index j, then the function

{±1}k−1 → PT(c) ,

(ε2, . . . , εk) 7→ pt(ζn11 · ζ

ε2 · n22 · · · ζεk · nkk )

is bijective. Here ζj is the number de�ned in formula (4.7) for the prime pj .

2. Otherwise, the set PT(c) is empty.



Exercise 6.2. Prove Theorem 6.1.

(Hint: use Theorem 5.2, and the symmetries in Figure 2.3.)

Here is an immediate consequence of Theorem 6.1.

Corollary 6.3. Let c be an integer > 1, with prime decomposition as inTheorem 6.1.

1. If pj ≡ 1 mod 4 for every index j, then the number of reduced orderedPythagorean triples with hypotenuse c is 2k−1.

2. Otherwise, there are no reduced ordered Pythagorean triples withhypotenuse c.

As I said before, this fact is not new; see discussion athttp://mathworld.wolfram.com/PythagoreanTriple.html.

But our proof is possibly new.Amnon Yekutieli (BGU) Pythagorean Triples 27 / 28

http://mathworld.wolfram.com/PythagoreanTriple.html


Theorem 6.1 is constructive. It lets us solve the next exercise easily.

Exercise 6.4.

1. Find the two reduced ordered Pythagorean triples with hypotenuse 85.2. Find the only reduced ordered Pythagorean triple with hypotenuse 289.

∼ END ∼


Pythagorean Triples, Complex Numbers, Abelian ... - BGU Mathamyekut/lectures/pythagoras/notes.pdf · (4.3) a = u Y1 i=1 qni i; with u 2T, n i 2Z, and all but ˙nitely many n i are

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