Download Pythagorean Triples, Complex Numbers, Abelian Groups and Prime

1. Pythagorean Triples
1. Pythagorean Triples
Pythagorean Triples, Complex Numbers, Abelian
Groups and Prime Numbers
A Pythagorean triple is a triple (a, b, c) of positive integers, satisfying
a2 + b 2 = c 2 .
(1.1)
Amnon Yekutieli
The reason for the name is, of course, because these are the sides of a right
angled triangle:
Department of Mathematics
Ben Gurion University
email: [email protected]
Notes available at
http://www.math.bgu.ac.il/~amyekut/lectures
c
b
written 7 June 2015
a
Amnon Yekutieli (BGU)
Pythagorean Triples
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Amnon Yekutieli (BGU)
1. Pythagorean Triples
We say that the triples (a, b, c) and (a′ , b′ , c ′ ) are equivalent if the
corresponding triangles are similar.
′
′
′
′
′
Question 1.3. Are there infinitely many reduced ordered Pythagorean
triples?
(a , b , c ) = (ra, rb, rc)
The answer is yes. This was already known to the ancient Greeks. There is a
formula attributed to Euclid for presenting all Pythagorean triples, and it
proves that there infinitely many reduced ordered triples.
or
(a , b , c ) = (rb, ra, rc).
Clearly r is rational.
This formula is somewhat clumsy, and I will not display it. It can be found
on the web, e.g. here:
We say that the triple (a, b, c) is reduced if the greatest common divisor of
these numbers is 1. The triple is called ordered if a ≤ b.
http://en.wikipedia.org/wiki/Pythagorean_triple
It is easy to see that any triple (a, b, c) is equivalent to exactly one reduced
ordered triple (a′ , b′ , c ′ ).
http://mathworld.wolfram.com/PythagoreanTriple.html
Later we will see a nice geometric argument, essentially based on the Euclid
formula, to show there infinitely many reduced ordered triples.
Exercise 1.2. Let (a, b, c) be a reduced ordered triple. Then c is odd, and
a < b.
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Here is an interesting question:
This means that there is a positive number r, such that
′
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1. Pythagorean Triples
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Amnon Yekutieli (BGU)
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1. Pythagorean Triples
2. Complex Numbers on the Unit Circle
For a given integer c > 1, let us denote by PT(c) the set of all reduced ordered
Pythagorean triples with hypotenuse c.
2. Complex Numbers on the Unit Circle
A more interesting question is this:
Question 1.4. What is the size of the set PT(c) ?
It was observed a long time ago that Pythagorean triples can be encoded as
complex numbers on the unit circle.
The answer to this was found in the 19th century. We will see it at the end of
the lecture.
Starting from a reduced ordered Pythagorean triple (a, b, c), we pass to the
complex number
z := a + b · i,
An even more interesting question is this:
that has absolute value c.
Question 1.5. Given c, is there an effective way to find the reduced
ordered Pythagorean triples with hypotenuse c ?
Consider the complex number
An effective method (possibly new!) will be presented at the end of the talk.
Amnon Yekutieli (BGU)
Pythagorean Triples
ζ = s + r · i :=
(2.1)
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Amnon Yekutieli (BGU)
2. Complex Numbers on the Unit Circle
a b
z
= + · i.
|z|
c
c
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2. Complex Numbers on the Unit Circle
The number ζ has rational coordinates, it is on the unit circle, in the second
octant, and is different from i.
Actually, there are 8 different numbers on the unit circle that encode the
same Pythagorean triple:
± ζ, ±i · ζ, ±ζ̄, ±i · ζ̄.
(2.2)
z
c
b
a
We can recover the number z, and thus the reduced ordered Pythagorean
triple (a, b, c), by clearing the denominators from the pair of rational
numbers (r, s) = ( ab , ac ).
Amnon Yekutieli (BGU)
Pythagorean Triples
Figure 2.3.
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2. Complex Numbers on the Unit Circle
2. Complex Numbers on the Unit Circle
Here is a geometric proof of the proposition.
Given a complex number ζ with rational coordinates on the unit circle, other
than the four special points ±1, ±i, let us denote by pt(ζ) the unique reduced
ordered Pythagorean triple (a, b, c) that ζ encodes.
In this fashion we obtain a function pt from the set of complex numbers on
the unit circle with rational coordinates (not including the four special
points), to the set of reduced ordered Pythagorean triples.
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 on
the straight line connecting i and ζ.
This function is surjective, and it is 8 to 1.
Therefore, to show that there are infinitely many reduced ordered
Pythagorean triples, it suffices to prove:
Proposition 2.4. There are infinitely many complex numbers on the unit
circle with rational coordinates.
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Amnon Yekutieli (BGU)
2. Complex Numbers on the Unit Circle
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3. The Circle as an Abelian Group
3. The Circle as an Abelian Group
Exercise 2.5. Show that ζ has rational coordinates iff the number f (ζ) is
rational.
(Hint: use similar triangles.)
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
Since there are infinitely many rational numbers, we are done.
G(R) := S1 = {ζ ∈ C | |ζ| = 1}.
The set G(R) is a group under complex multiplication, because
|ζ1 · ζ2 | = |ζ1 | · |ζ2 |
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Amnon Yekutieli (BGU)
and |ζ −1 | = |ζ|−1 .
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3. The Circle as an Abelian Group
3. The Circle as an Abelian Group
We first locate all the elements of finite order in the group G(Q). These are the
roots of 1, namely the elements ζ satisfying ζ n = 1 for some positive integer
n.
Let G(Q) be the subset of G(R) consisting of points with rational
coordinates; namely
(3.1)
Algebraic number theory tells us that there are just four of them:
G(Q) = {ζ = s + r · i | s, r ∈ Q, s2 + r 2 = 1}.
1, i, −1, −i.
(3.3)
Exercise 3.2. Prove that G(Q) is a subgroup of G(R).
Recall that to answer Question 1, namely to show there are infinitely many
reduced ordered Pythagorean triples, it suffices to prove that the abelian
group G(Q) is infinite.
Thus, if we take any element ζ ∈ G(Q) other than those four numbers, the
cyclic subgroup that it generates
{ζ n | n ∈ Z} ⊂ G(Q)
will be infinite!
Amnon Yekutieli (BGU)
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Amnon Yekutieli (BGU)
3. The Circle as an Abelian Group
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3. The Circle as an Abelian Group
Let us consider the familiar reduced ordered Pythagorean triple (3, 4, 5).
The corresponding number in G(Q) is
ζ :=
3
5
+ 45 · i,
Exercise 3.4. Find a reduced Pythagorean triple with hypotenuse c = 3125.
and it is not one of the four special numbers in (3.3). So this element has
infinite order in the group G(Q).
(Later we will see that there is only one!)
Here are the first positive powers of ζ, and the corresponding triples.
ζn
n
1
Amnon Yekutieli (BGU)
3
5
+
Remark: the algebraic number theory used above, and all that is needed to
complete the proofs in this lecture, can be found in the book
pt(ζ n ) = (an , bn , cn )
4
5i
“Algebra”, by M. Artin, Prentice-Hall.
(3, 4, 5)
2
7
+
− 25
24
25 i
(7, 24, 25)
3
117
− 125
+
44
125 i
(44, 117, 125)
4
527
−
− 625
336
625 i
(336, 527, 625)
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4. The Ring of Gauss Integers
4. The Ring of Gauss Integers
The group
of unity
4. The Ring of Gauss Integers
A×
of invertible elements of A turns out to be the group of roots
T := {±1, ±i}.
(4.2)
Consider the ring of Gauss integers
A := Z[i] = {m + n · i | m, n ∈ Z}.
Unique factorization tells us that any element a ∈ K × can be written
uniquely a product
Let us denote its field of fractions by
K := Q[i] = {s + r · i | s, r ∈ Q}.
a = u·
(4.3)
∞
Y
qini ,
i=1
The reason we want to look at K is this: equation (3.1) shows that
with u ∈ T , ni ∈ Z, and all but finitely many ni are 0 (so the product is
actually finite).
G(Q) = {ζ ∈ K | |ζ| = 1}.
(4.1)
It is known that the ring A is a unique factorization domain.
We see that as an abelian group,
There are countably many primes in A. Let us enumerate them as
(4.4)
q1 , q2 , q3 , . . . .
Amnon Yekutieli (BGU)
Pythagorean Triples
K × = T × F,
where F is the free abelian group with basis {qi }i=1,2,... .
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Amnon Yekutieli (BGU)
4. The Ring of Gauss Integers
Pythagorean Triples
It is known that the primes of the ring A = Z[i] are of three kinds.
Finally, let p be a prime of Z satisfying
The first kind is the prime
(4.5)
q := 1 + i.
p ≡ 1 mod 4.
For example p = 5 or p = 13.
It is the only prime divisor of 2 in A, with multiplicity 2 :
Then there are two primes q and q̄ in A, conjugate to each other but not
equivalent (i.e. q̄ ∈
/ T · q), such that
q2 = (1 + i)2 = 1 + i2 + 2 · i = i · 2.
Next let p be a prime in Z satisfying
p = u · q · q̄
(4.6)
p ≡ 3 mod 4.
for some u ∈ T .
For example p = 3 or p = 7.
The numbers q, q̄ are the third kind of primes of A.
Then p is also prime in A. This is the second kind of primes.
These will be the interesting primes for us.
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4. The Ring of Gauss Integers
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4. The Ring of Gauss Integers
5. The Group Structure of the Rational Circle
It is not hard to find the decomposition (4.6). The number p is a sum of two
squares in Z :
p = m2 + n2 .
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:
We then take
q := m + n · i
and q̄ := m − n · i.
p1′ := 5 ,
p2′ := 13 ,
Each such prime pj′ of Z has a prime decomposition in the ring A = Z[i] :
Consider the number
pj′ = uj · qj′ · qj′ .
ζ := q/q̄ ∈ K.
(4.7)
p3′ := 17 , . . . .
We use it to define the element
It has absolute value 1, and hence, by (4.1), it belongs to the group G(Q).
ζj′ := qj′ / qj′ ∈ G(Q).
(5.1)
Thus we get a sequence of elements {ζj′ }j=1,2,... in the group G(Q).
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5. The Group Structure of the Rational Circle
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5. The Group Structure of the Rational Circle
Here is our main result.
Sketch of Proof.
Theorem 5.2.
Consider the prime decomposition of ζ as an element of K × , as in (4.3).
Any element ζ ∈ G(Q) is uniquely a product
ζ = u·
∞
Y
For each prime qi of A we calculate its absolute value |qi | ∈ R.
The equality
ζj′ nj ,
j=1
1=
implies that the primes qi that do not come in pairs must have multiplicity
ni = 0.
In other words, the abelian group G(Q) is a product
The primes that do come in pairs, namely the primes of the third kind, must
have opposite multiplicities. Thus they appears as powers of the
corresponding number ζ.
G(Q) = T × F ′ ,
where F ′ is the free abelian group with basis {ζj′ }j=1,2,... .
Pythagorean Triples
|qi |ni
i=1
with u ∈ T , nj ∈ Z, and all but finitely many of the nj are 0.
Amnon Yekutieli (BGU)
∞
Y
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6. Back to Pythagorean Triples
6. Back to Pythagorean Triples
Theorem 6.1. Let c be an integer greater than 1, with prime decomposition
c = p1n1 · · · pknk
in Z. Here p1 < · · · < pk are positive primes; n1 , . . . , nk are positive integers;
and k is a positive integer.
6. Back to Pythagorean Triples
1. If pj ≡ 1 mod 4 for every index j, then the function
Recall that for a number ζ ∈ G(Q) − T , we write pt(ζ) for the
corresponding reduced ordered Pythagorean triple.
{±1}k−1 → PT(c) ,
Here is our explicit presentation of all reduced ordered Pythagorean triples.
(ǫ2 , . . . , ǫk ) 7→ pt(ζ1n1 · ζ2ǫ2 · n2 · · · ζkǫk · nk )
is bijective. Here ζj is the number defined in formula (4.7) for the prime pj .
2. Otherwise, the set PT(c) is empty.
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Amnon Yekutieli (BGU)
6. Back to Pythagorean Triples
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6. Back to Pythagorean Triples
Exercise 6.2. Prove Theorem 6.1.
(Hint: use Theorem 5.2, and the symmetries in Figure 2.3.)
Theorem 6.1 is constructive. It lets us solve the next exercise easily.
Here is an immediate consequence of Theorem 6.1.
Exercise 6.4.
Corollary 6.3. Let c be an integer > 1, with prime decomposition as in
Theorem 6.1.
1. Find the two reduced ordered Pythagorean triples with hypotenuse 85.
1. If pj ≡ 1 mod 4 for every index j, then the number of reduced ordered
Pythagorean triples with hypotenuse c is 2k−1 .
2. Find the only reduced ordered Pythagorean triple with hypotenuse 289.
2. Otherwise, there are no reduced ordered Pythagorean triples with
hypotenuse c.
∼ END ∼
As I said before, this fact is not new; see discussion at
http://mathworld.wolfram.com/PythagoreanTriple.html.
But our proof is possibly new.
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