EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 3, 2024, 2127-2141
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Structure of Primitive Pythagorean Triples in
Generating Trees

Lucia Kőszegyová 1,∗, Evelin Csókási1, Juraj Hirjak1

1 Institute of Mathematics, Faculty of Science, P.J. Šafárik University in Košice, Košice,
Slovakia

Abstract. A Pythagorean triple is a triple of positive integers (a, b, c) such that a2 + b2 = c2. If
a, b are coprime, then it is called a primitive Pythagorean triple. It is known that every primitive
Pythagorean triple can be generated from the triple (3, 4, 5) using multiplication by unique number
and order of three specific 3 × 3 matrices, which yields a ternary tree of triplets. Two such trees
were described by Berggren and Price, respectively. A different approach is to view the primitive
Pythagorean triples as points in the three-dimensional Euclidean space. In this paper, we prove
that the triple of descendants of any primitive Pythagorean triple in Berggren’s or Price’s tree forms
a triangle (and therefore defines a plane), and we present our results related to these triangles (and
these planes).

2020 Mathematics Subject Classifications: 11D09, 11A41, 11C99

Key Words and Phrases: Primitive Pythagorean triples, Berggren’s tree, Price’s tree, Euclid’s
formula

1. Introduction

Primitive Pythagorean triples are an interesting object of the number theory, and
they have potential to be also used while studying the properties in other areas of the
mathematics (e.g. in geometry, or in graph theory). The description and knowledge of
their properties can also be used practically, for example in cryptography, considering
that there are infinitely many primitive Pythagorean triples, and the fact that such triples
have the ability of creating probability events [15]. Moreover, sufficiently large primitive
pythagorean triples can be used as keys by certification authority - sending two numbers to
two parties as seeds to generate the key, and keeping the third number for audit purposes
[5].

To use primitive pythagorean triples, we need to understand them and their generating
better. Currently, there are various ways of generating integer solutions of Pythagorean

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5323

Email addresses: lucia.janickova@upjs.sk (L. Kőszegyová),
csokasi.evelinke@gmail.com (E. Csókási), juraj.hirjak@student.upjs.sk (J. Hirjak)

https://www.ejpam.com 2127 © 2024 EJPAM All rights reserved.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2128

equation a2 + b2 = c2. Euclid’s formula, Berggren’s tree and Price’s tree are among
the best known ones, however there are many others. Some examples can be found e.g.
in [13]. The matrices used to generate Berggren’s tree produce solutions (a, b, c) with
coprime elements, i.e. the primitive Pythagorean triples, which can be viewed as lengths
of the sides of Pythagorean triangles. Another methods of gerating these triples use
sequences, generating the following triples from a starting triple. One of the newest of
these appraoches is described by [5]. Other approaches to primitive Pythagorean triples
were introduced e.g. by [2].

Some properties of the Pythagorean triangles were already described. E.g., the number
of primitive Pythagorean triples with a given inradius [11], triples with common lengths
of leg [6] or height of primitive Pythagorean triples (the difference between length of the
hypotenuse and length of even leg) [4]. Further, it was described how to find the pairs
of Pythagorean triangles with equal areas, or primitive Pythagorean triangles having the
same perimeter [1]. Also, some numerical properties of primitive Pythagorean triples were
studied e.g. in [3], [8].

The primitive Pythagorean triples can be also viewed as coordinates of the points in
the three-dimensional Euclidean space. In this paper, we consider the triples which are
generated from a given triple in either Berggren’s or Price’s tree of primitive Pythagorean
triples (such trees always produce three descendant triples from a given triple). Then
we assign these descendant triples the vertices in three-dimensional space and prove that
these points are not colinear, therefore they form a triangle (or a plane).

We focus on these triangles formed by the descendant triples in either Berggren’s
or Price’s tree of primitive Pythagorean triples, and describe some of their properties.
Namely we study the planes containing these triangles; and the types of these triangles.

2. Preliminary

First, we present some basic notations and definitions which we use through the paper.
Let us denote N := {1, 2, 3, . . . } and N0 := N ∪ {0}.
For a matrix M , denote its transpose as M⊤.
For a vector u⃗, we denote its norm as |u⃗|.

Definition 1. A triple of positive integers (a, b, c) is called a Pythagorean triple if a2+b2 =
c2. Moreover, if gcd(a, b) = 1, then we call (a, b, c) a primitive Pythagorean triple.

Definition 2. If (a, b, c) is a Pythagorean triple, we say that a triangle corresponds to this
triple if its sides have lengths a, b, c. A triangle corresponding to a (primitive) Pythagorean
triple is called a (primitive) Pythagorean triangle.

For convenience, we use the abbreviation PPT for both primitive Pythagorean triple
and primitive Pythagorean triangle.

It is easy to see the parity of the individual components of a PPT, see e.g. [8]:

Lemma 1. If (a, b, c) is a primitive Pythagorean triple, then one of a, b is even and the
other one is odd, and c is odd.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2129

In the following sections, we consider generating the primitive Pythagorean triples
using the trees described by Berggren and Price, respectively. In these trees, the compo-
nents of PPT (a, b, c) are ordered such that a is odd and b is even. We use this order of
components in the rest of our article.

In the proofs concerning PPTs, it is often useful to express the primitive Pythagorean
triples using Euclid’s formula [10]:

Theorem 1 (Euclid’s formula). Triple (a, b, c) is a primitive Pythagorean triple with odd
a if and only if there exist m,n ∈ N of different parity, such that gcd(m,n) = 1, m > n
and

a = m2 − n2, b = 2mn, c = n2 +m2. (1)

3. Descendants in Berggren’s tree

According to Berggren, every primitive Pythagorean triple (a, b, c) with odd a can be
generated from the triple (3, 4, 5) by unique three-fold ascent using the three matrices
U,A,D [7]:

U =

1 −2 2
2 −1 2
2 −2 3

 , A =

1 2 2
2 1 2
2 2 3

 , D =

−1 2 2
−2 1 2
−2 2 3

 .

Further, it is easy to show that for every n ∈ N the following holds:

Un =

 1 −2n 2n
2n 1− 2n2 2n2

2n −2n2 2n2 + 1

 ,

An =

 (−1)n

2 + a1
(−1)n+1

2 + a1 a2
(−1)n+1

2 + a1
(−1)n

2 + a1 a2
a2 a2 2a1

 ,

Dn =

1− 2n2 2n 2n2

−2n 1 2n
−2n2 2n 2n2 + 1

 ,

where a1 =
1
4 [(3− 2

√
2)n + (3 + 2

√
2)n], a2 = −(3−2

√
2)

n

2
√
2

+
(3+2

√
2)

n

2
√
2

.

Theorem 2. If (a, b, c) is a primitive Pythagorean triple with odd a, and M is a matrix
such that M ∈ {U,A,D}, then M · (a, b, c)⊤ is a primitive Pythagorean triple with an odd
first component.

Berggren’s tree of PPT contains all PPTs, and each PPT is generated by a unique
sequence of matrix multiplication [7]. The first few levels of Berggren’s tree can be seen
in the Figure 1.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2130

(3, 4, 5)

(5, 12 ,13)

(21, 20, 29)

(15, 8, 17)

(7, 24, 25)

(55, 48, 73)

(45, 28, 53)

(39, 80, 89)

(119, 120, 169)

(77, 36, 85)

(33, 56, 65)

(65, 72, 97)

(35, 12, 37)

U

A

D

U

A

D

U

A

D

U

A

D

Figure 1: Berggren’s tree of PPTs.

Definition 3. Let P be a primitive Pythagorean triple. We say that P is the parent of
the triples UP⊤, AP⊤, DP⊤. The triples UP⊤, AP⊤, DP⊤ are called the descendants of
P in Berggren’s tree.

Proposition 1. Let P be a primitive Pythagorean triple. Then the points with the coor-
dinates UP⊤, AP⊤, DP⊤ are not collinear.

Proof. Let P = (a, b, c). We consider the vectors u⃗ = AP⊤ − UP⊤ and v⃗ = DP⊤ −
UP⊤. It is easy to show that

u⃗ = (A− U)P⊤ =

0 4 0
0 2 0
0 4 0

 ·

a
b
c

 =

4b
2b
4b

 ,

v⃗ = (D − U)P⊤ =

−2 4 0
−4 2 0
−4 4 0

 ·

a
b
c

 =

−2a+ 4b
−4a+ 2b
−4a+ 4b

 .

By way of contradiction, assume that these vectors are linearly dependent, i.e., that
there exists nonzero real number k such that u⃗ = kv⃗. Then (4b, 2b, 4b) = k(−2a+4b,−4a+
2b,−4a+ 4b), hence 4b = k(−2a+ 4b) and 4b = k(−4a+ 4b). This implies that

k(−2a+ 4b) = k(−4a+ 4b)

−2a+ 4b = −4a+ 4b

2a = 0

which is a contradiction with a ∈ N. Therefore, u⃗, v⃗ are linearly independent and the
points UP⊤, AP⊤, DP⊤ are not collinear.

This implies that the points with the coordinates UP⊤, AP⊤, DP⊤ determine a plane,
and a triangle. In the following, we prove some properties of this plane (and this triangle).
For simplicity, we use the following notion:

Definition 4. Let P be a primitive Pythagorean triple and let UP⊤, AP⊤, DP⊤ be its
descendants in Berggren’s tree. The triangle with vertices with the coordinates UP⊤, AP⊤,
DP⊤ is called the descendant triangle of P in Berggren’s tree, and we denote it by △B(P ).



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2131

Proposition 2. Let P = (a, b, c) be a primitive Pythagorean triple. Then the triangle
△B(P ) belongs to the plane

2x+ 2y − 3z + c = 0. (2)

Proof. According to the Proposition 1, vertices UP⊤, AP⊤, DP⊤ form a triangle,
therefore, they define a plane. Analogously to the previous proof, we consider vectors
u⃗ = AP⊤ − UP⊤ = (4b, 2b, 4b) and v⃗ = DP⊤ − UP⊤ = (−2a+ 4b,−4a+ 2b,−4a+ 4b).

Firstly, we compute the normal vector of the wanted plane as cross product of these
vectors:

u⃗× v⃗ = (8ab, 8ab,−12ab) ≈ (2, 2,−3).

Therefore, 2x + 2y − 3z + d = 0 is the equation of wanted plane for some d ∈ R. Since
UP⊤ belongs to this plane, we get

2(a− 2b+ 2c) + 2(2a− b+ 2c)− 3(2a− 2b+ 3c) + d = 0

which yields d = c. Hence, the wanted plane is 2x+ 2y − 3z + c = 0.

It directly follows that:

Corollary 1. If P,Q are primitive Pythagorean triples, then the planes defined by triangles
△B(P ) and △B(Q) are parallel.

Corollary 2. Each primitive Pythagorean triple belongs to a plane 2x+ 2y − 3z + d = 0
for some odd d ∈ N.

Proof. According to [7], Berggren’s tree contains all primitive Pythagorean triples.
Therefore, if P is a PPT, then it is either (3, 4, 5) or it is a descendant triple of some triple
(e, f, g) in Berggren’s tree such that g ∈ N is odd.

If P = (3, 4, 5) then it clearly belongs to the plane 2x+ 2y − 3z + 1 = 0.
If P is a descendant triple of some triple (e, f, g) in Berggren’s tree, then by Proposition

2, it belongs to the plane 2x+ 2y − 3z + g = 0 where g ∈ N is odd.

In [14], Tripathi proved the following Lemma:

Lemma 2. For odd n ∈ N, the number of primitive Pythagorean triples (a, b, n) is

P∗
2 (n) =

{
2ω(n)−1, if n ≥ 3 and no prime of the form 4k + 3 divides n,
0, if n = 1 or n has a prime divisor of the form 4k + 3,

where ω(n) is the number of prime divisors of n.

We use Lemmas 1 and 2 to determine the number of descendant triangles in the
individual planes.

Proposition 3. If c ∈ N is odd, then the plane 2x + 2y − 3z + c = 0 contains P∗
2 (c)

descendant triangles △B.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2132

Proof. According to Lemma 2, for each odd c ∈ N, there is P∗
2 (c) primitive Pythagorean

triples P with the third component c. Further, the Proposition 2 yields that for each such
P , the descendant triangle △B(P ) belongs to the plane 2x+ 2y − 3z + c = 0.

It is easy to see that the plane 2x+ 2y − 3z + c = 0 cannot contain more descendant
triangles. Assume that there is a descendant triangle △B(Q) in 2x+2y− 3z+ c = 0 such
that Q = (a, b, n), n ̸= c. However, by the Proposition 2, △B(Q) also belongs to the plane
2x+ 2y − 3z + n = 0 which yields n = c, a contradiction.

Corollary 3. For each odd integer n ≥ 3, the plane 2x+2y−3z+n = 0 contains 3·P∗
2 (n)

primitive Pythagorean triples. The plane 2x+ 2y − 3z + 1 = 0 contains a single primitive
Pythagorean triple, (3, 4, 5).

Proof. Let n = 1. If (a, b, c) is a primitive Pythagorean triple belonging to 2x+ 2y −
3z + 1 = 0, then by Euclid’s formula we get

2(m2 − n2) + 2(2mn)− 3(m2 + n2) + 1 = 0

m2 − 4mn+ 5n2 − 1 = 0

Assuming that m is a variable and n is a parameter, we get

m =
4n±

√
4− 4n2

2

According to Euclid’s formula, n,m ∈ N, which yields 4− 4n2 ≥ 0 ⇐⇒ n = 1, therefore,
m = 2. Hence, a primitive Pythagorean triple (a, b, c) belongs to the plane 2x+2y−3z+1 =
0 ⇐⇒ [(a, b, c) = (m2 − n2, 2mn,m2 + n2) where n = 1 and m = 2] ⇐⇒ (a, b, c) =
(3, 4, 5).

Let n be an odd integer, n ≥ 3. By Proposition 3, the plane 2x + 2y − 3z + n = 0
contains P∗

2 (n) descendant triangles △B. Each vertex of △B is a PPT, hence there are
at least 3 · P∗

2 (n) PPTs in the plane 2x+ 2y − 3z + n = 0.
By way of contradiction, we show that there are no more PPTs in 2x+2y−3z+n = 0.

Let us assume that there is a PPT (a, b, c) in 2x + 2y − 3z + n = 0 such that (a, b, c)
is not a descendant of any PPT with the third component n in Berggren’s tree. Then
(a, b, c) = (3, 4, 5) or (a, b, c) is a descendant of some PPT (d, e, f) such that f ̸= n.

If (a, b, c) = (3, 4, 5), then from (a, b, c) belonging to 2x+2y−3z+n = 0, we get n = 1,
a contradiction with n ≥ 3.

If (a, b, c) is a descendant of some PPT (d, e, f), such that f ̸= n, then by Proposition
2, (a, b, c) belongs to the plane 2x + 2y − 3z + f = 0. However, the triple (a, b, c) also
belongs to the plane 2x+ 2y − 3z + n = 0, which yields f = n, a contradiction.

A primitive Pythagorean triple can be viewed as a right triangle, and the points corre-
sponding to the descendants of a PPT in Berggren’s tree also form a triangle. It is natural
to ask whether the descendant triangle △B(P ) can also be a right triangle for some P .
The following proposition offers the answer to this question.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2133

Proposition 4. Let P be a primitive Pythagorean triple. The triangle △B(P ) is a non-
right triangle.

Proof. Let P = (a, b, c) and let the vectors u⃗, v⃗, w⃗ be as follows:

u⃗ = AP⊤ − UP⊤,

v⃗ = DP⊤ − UP⊤,

w⃗ = DP⊤ −AP⊤.

To show that the triangle with the vertices UP⊤, AP⊤, DP⊤ is a non-right triangle, it
is sufficient to show that u⃗ · v⃗ ̸= 0, u⃗ ·w⃗ ̸= 0, v⃗ ·w⃗ ̸= 0. According to the proof of Proposition
1, u⃗ = (4b, 2b, 4b) and v⃗ = (−2a+ 4b,−4a+ 2b,−4a+ 4b). Analogously,

w⃗ = (D −A)P⊤ =

−2 0 0
−4 0 0
−4 0 0

 ·

a
b
c

 =

−2a
−4a
−4a

 .

1.) By way of contradiction, assume that u⃗ · v⃗ = 0. Clearly, u⃗ · v⃗ = (4b, 2b, 4b) · (−2a+
4b,−4a + 2c,−4a + 4b) = −32ab + 36b2. Since a, b > 0, we get b = 8a

9 . However, (a, b, c)
satisfies the Pythagorean equation which yields

a2 +
64a2

81
= c2

√
145

a

9
= c

which is a contradiction with a, c ∈ N.
2.) Similarly, u⃗ · w⃗ = (4b, 2b, 4b) · (−2a,−4a,−4a) = −32ab and from a, b > 0 it follows

that u⃗ · w⃗ ̸= 0.
3.) Finally, v⃗ · w⃗ = (−2a+4b,−4a+2c,−4a+4b) · (−2a,−4a,−4a) = 36a2− 32ab and

we get a contradiction analogously to the case u⃗ · v⃗.

Also, it can be proven that the triangle △B(P ) is neither isosceles nor equilateral
triangle.

Proposition 5. Let P be a primitive Pythagorean triple. The triangle △B(P ) is not an
isosceles triangle.

Proof. Let P = (a, b, c). Similarly like above, we consider the vectors u⃗, v⃗, w⃗:

u⃗ = AP⊤ − UP⊤ = (4b, 2b, 4b),

v⃗ = DP⊤ − UP⊤ = (−2a+ 4b,−4a+ 2b,−4a+ 4b),

w⃗ = DP⊤ −AP⊤ = (−2a,−4a,−4a).

It remains to prove that each of these vectors has a different norm. Clearly,

|u⃗| =
√
36b2,



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2134

|v⃗| =
√

36a2 − 64ab+ 36b2,

|w⃗| =
√
36a2.

1.) By way of contradiction, let us assume that |u⃗| = |w⃗|. Then
√
36b2 =

√
36a2 =⇒ 36b2 = 36a2 =⇒ a = b

which is a contradiction with (a, b, c) being a PPT.
2.) If |u⃗| = |v⃗|, then

√
36b2 =

√
36a2 − 64ab+ 36b2

36b2 = 36a2 − 64ab+ 36b2

0 = 36a2 − 64ab

0 = 9a− 16b.

Applying Euclid’s formula, we get

0 = 9(m2 − n2)− 16(2mn)

16(2mn) = 9(m2 − n2)

where m,n have different parity. Then 16(2mn) is clearly even and 9(m2 − n2) is odd,
which is a contradiction.

3.) Finally, let |w⃗| = |v⃗|. Then
√
36a2 =

√
36a2 − 64ab+ 36b2

36a2 = 36a2 − 64ab+ 36b2

0 = −16a+ 9b

Applying Euclid’s formula, we get

0 = −16(m2 − n2) + 9(2mn)

0 = −8m2 + 8n2 + 9mn.

Therefore

n =
−9m±

√
337m

16
̸∈ N,

which is a contradiction with n ∈ N.

Corollary 4. Let P be a primitive Pythagorean triple. The triangle △B(P ) is not an
equilateral triangle.

Further, we study the area, inradius and the radius of the circumcircle of the descendant
triangles in Berggren’s tree and present the related results.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2135

Proposition 6. Let P = (a, b, c) be a primitive Pythagorean triple. The triangle △B(P )
has the area

S = 2
√
17ab. (3)

For proof, see [9].

Corollary 5. Let P,Q be primitive Pythagorean triples. Then the triangles corresponding
to P and Q have the same area if and only if the triangles △B(P ) and △B(Q) have the
same area.

It would be interesting to know whether there even exist any PPTs with the same area.
According to [13], the smallest PPTs with the same area are (20, 21, 29) and (12, 35, 37).
Both of them have the area 210. Further, the following theorem was proved in [13]:

Theorem 3. For every n ∈ N, there exist n primitive Pythagorean triples with different
third components and the same area.

4. Descendants in Price’s tree

Another well known tree of primitive Pythagorean triples was described by Price in
[12]. Analogously like in Berggren’s tree, Price proved that each PPT can be generated
from (3, 4, 5) by unique three-fold ascent using the three matrices M1,M2,M3 :

M1 =

 2 1 −1
−2 2 2
−2 1 3

 , M2 =

2 1 1
2 −2 2
2 −1 3

 , M3 =

2 −1 1
2 2 2
2 1 3

 .

It is easy to show that

Mn
1 =

 2n −2n + 1 2n − 1
22n − 2n 2n 22n − 2n

22n − 2n 2n − 1 22n − 2n + 1

 ,

Mn
2 =


22n+2+(−1)n·2n+4

9
(−1)n+1·2n+1

3
22n+2+(−1)n·2n−5

9
22n+(−1)n·2n

3 (−1)n · 2n 22n+(−1)n·2n
3

5·22n+(−1)n·2n−4
9

(−1)n·2n−1
3

5·22n+(−1)n·2n+5
9

 ,

Mn
3 =

 2n 2n − 1 −2n + 1
−22n + 2n 2n 22n − 2n

−22n + 2n 2n − 1 22n − 2n + 1

 .

Theorem 4. If (a, b, c) is a primitive Pythagorean triple with odd a, and M is a matrix
such that M ∈ {M1,M2,M3}, then M · (a, b, c)⊤ is a primitive Pythagorean triple with an
odd first component.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2136

(3, 4, 5)

(5, 12 ,13)

(15, 8, 17)

(7, 24, 25)

(9, 40, 41)

(35, 12 ,37)

(11, 60, 61)

(21, 20, 29)

(55, 48, 73)

(39, 80, 89)

(13, 84, 85)

(63, 16, 65)

(15, 112, 113)

M1

M2

M3

M1

M2

M3

M1

M2

M3

M1

M2

M3

Figure 2: Price’s tree of PPTs.

Price’s tree of PPT contains all PPTs, and each PPT is generated by a unique sequence
of matrix multiplication [12]. The first levels of Price’s tree are in the Figure 2.

Similar relations that hold in Berggren’s tree of PPTs can be also proved in Price’s
tree. We will use the analogous terms and notations.

Definition 5. Let R be a primitive Pythagorean triple. We say that R is the parent of the
triples M1R

⊤,M2R
⊤,M3R

⊤. The triples M1R
⊤,M2R

⊤,M3R
⊤ are called the descendants

of R in Price’s tree.

Proposition 7. Let R be a primitive Pythagorean triple. Then the points with the coor-
dinates M1R

⊤,M2R
⊤,M3R

⊤ are not collinear.

Proof. Let R = (a, b, c). We consider the vectors p⃗ = M2R
⊤ − M1R

⊤ and q⃗ =
M3R

⊤ −M1R
⊤. It is easy to show that

p⃗ = (M2 −M1)R
⊤ =

0 0 2
4 −4 0
4 −2 0

 ·

a
b
c

 =

 2c
4a− 4b
4a− 2b

 ,

q⃗ = (M3 −M1)R
⊤ =

0 −2 2
4 0 0
4 0 0

 ·

a
b
c

 =

−2b+ 2c
4a
4a

 .

By way of contradiction, assume that these vectors are linearly dependent, i.e., that
there exists nonzero real number k such that (2c, 4a − 4b, 4a − 2b) = k(−2b + 2c, 4a, 4a),
hence 4a− 4b = k(4a) and 4a− 2b = k(4a). This implies

4a− 4b = 4a− 2b

−4b = −2b

b = 0

which is a contradiction with b ∈ N. Therefore, p⃗, q⃗ are linearly independent and the
points M1R

⊤,M2R
⊤,M3R

⊤ are not collinear.

Therefore, the points M1R
⊤,M2R

⊤,M3R
⊤ determine a plane, and a triangle. In the

following, we prove some properties of this plane (and this triangle).



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2137

Definition 6. Let R be a primitive Pythagorean triple and let M1R
⊤,M2R

⊤, M3R
⊤ be its

descendants in Price’s tree. The triangle with vertices with the coordinates M1R
⊤,M2R

⊤,M3R
⊤

is called the descendant triangle of R in Price’s tree, and we denote it by △P (R).

Proposition 8. Let R = (a, b, c) be a primitive Pythagorean triple. Then the triangle
△P (R) belongs to the plane

(2a)x+ (2a− b+ c)y + (−2a+ 2b− 2c)z + 2ab− 2ac+ 4bc− 4b2 = 0. (4)

Proof. According to the Proposition 7, the vertices M1R
⊤,M2R

⊤,M3R
⊤ form a tri-

angle, therefore, they define a plane. We consider the vectors p⃗ = (M2 − M1)R
⊤ =

(2c, 4a− 4b, 4a− 2b) and q⃗ = (M3 −M1)R
⊤ = (−2b+2c, 4a, 4a) and we compute the nor-

mal vector of the wanted plane as cross product of these vectors: p⃗× q⃗ = (−8ab,−8ab+
4b2 − 4bc, 8ab+ 8bc− 8b2) ≈ (2a, 2a− b+ c,−2a+ 2b− 2c).

Hence, (2a)x+(2a− b+ c)y+(−2a+2b−2c)z+d = 0 is the equation of wanted plane
for some d ∈ R. Since M1R

⊤ belongs to this plane, we get

(2a)(2a+ b− c) + (2a− b+ c)(−2a+ 2b+ 2c) + (−2a+ 2b− 2c)(−2a+ b+ 3c) + d = 0.

Moreover, c2 = a2 + b2, which yields d = 2ab− 2ac+4bc− 4b2. Therefore, △P (R) belongs
to the plane

(2a)x+ (2a− b+ c)y + (−2a+ 2b− 2c)z + 2ab− 2ac+ 4bc− 4b2 = 0.

Notice the difference between Price’s tree and Berggren’s tree. In Berggren’s tree, all
descendant triangles △B(P ) belong to planes parallel with the plane 2x + 2y − 3z = 0
(see the Proposition 2). In general, the results in Berggren’s tree are usually nicer than
analogous results in Price’s tree.

Also, similarly to Berggren’s tree, no plane defined by a descendant triangle △P (R) in
Price’s tree passes through the origin (0, 0, 0).

Proposition 9. Let (a, b, c) be a primitive Pythagorean triple. The plane (2a)x + (2a −
b + c)y + (−2a + 2b − 2c)z + 2ab − 2ac + 4bc − 4b2 = 0 does not pass through the origin
(0, 0, 0).

Proof. If (0, 0, 0) belongs to this plane, then 2ab − 2ac + 4bc − 4b2 = 0. After some
modifications, we get

ab− ac+ 2bc− 2b2 = 0

b(a− 2b)− c(a− 2b) = 0

(b− c)(a− 2b) = 0

which holds iff b − c = 0 or a − 2b = 0. However, b − c = 0 =⇒ b = c, a contradiction
with (a, b, c) being a PPT, and a − 2b = 0 =⇒ a = 2b which is a contradiction with a
being odd and 2b being even.

From the Propositions 4 and 5, we already know that in Berggren’s tree, the descendant
triangle △B(P ) fails to be either a right triangle or an isosceles triangle. The following
propositions offer the analogous results in Price’s tree.



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2138

Proposition 10. Let R be a primitive Pythagorean triple. The triangle △P (R) is a non-
right triangle.

Proof. Let us consider the vectors

p⃗ = M2R
⊤ −M1R

⊤ = (2c, 4a− 4b, 4a− 2b),

q⃗ = M3R
⊤ −M1R

⊤ = (−2b+ 2c, 4a, 4a),

r⃗ = M3R
⊤ −M2R

⊤ = (−2b, 4b, 2b).

To prove that △P (R) is a non-right triangle, it is sufficient to show that p⃗ · q⃗ ̸= 0, p⃗ · r⃗ ̸= 0
and q⃗ · r⃗ ̸= 0.

1.) By way of contradiction, let us assume that p⃗ · q⃗ = 0. This and a2 + b2 = c2 yield

(2c, 4a− 4b, 4a− 2b) · (−2b+ 2c, 4a, 4a) = 16a2 − 8b2 + 4c2 − 4cb = 0

20c2 − 24b2 − 4bc = 0

5c2 − 6b2 − bc = 0

6b2 + bc = 5c2.

In the beginning, we set the order of the components of (a, b, c) to be odd, even, odd.
Hence, there is an even natural number on the left side, and an odd natural number on
the right side, which is a contradiction.

2.) Analogously, let us assume that p⃗ · r⃗ = 0. Then

p⃗ · r⃗ = (2c, 4a− 4b, 4a− 2b) · (−2b, 4b, 2b) = 0

−20b2 + 24ab− 4bc = 0

5b− 6a+ c = 0.

Using Euclid’s formula, we get

5(2mn)− 6(m2 − n2) + (m2 + n2) = 0

10mn− 5m2 = −7n2

where m,n ∈ N have different parity. If m is even and n is odd, then the left side of
this equation is even and the right side is odd, which is a contradiction. If m is odd
and n is even, then the left side of this equation is odd and the right side is even, also a
contradiction. Hence, p⃗ · r⃗ ̸= 0.

3.) Finally, we assume that q⃗ · r⃗ = 0. Then

q⃗ · r⃗ = (−2b+ 2c, 4a, 4a) · (−2b, 4b, 2b) = 0

4b2 − 4bc+ 24ab = 0

b− c+ 6a = 0.

Using Euclid’s formula, we get

2mn− (m2 + n2) + 6(m2 − n2) = 0



L. Kőszegyová, E. Csókási, J. Hirjak / Eur. J. Pure Appl. Math, 17 (3) (2024), 2127-2141 2139

2mn+ 5m2 = 7n2

where m,n ∈ N have different parity. We get the contradiction analogously to the previous
case.

Proposition 11. Let R be a primitive Pythagorean triple. The triangle △P (R) is not an
isosceles triangle.

Proof. Let R = (a, b, c). Similarly like above, we consider the vectors p⃗, q⃗, r⃗:

p⃗ = M2R
⊤ −M1R

⊤ = (2c, 4a− 4b, 4a− 2b),

q⃗ = M3R
⊤ −M1R

⊤ = (−2b+ 2c, 4a, 4a),

r⃗ = M3R
⊤ −M2R

⊤ = (−2b, 4b, 2b).

Then

|p⃗| = 2
√
9a2 − 12ab+ 6b2

|q⃗| = 2
√

(c− b)2 + 8a2

|r⃗| = 2
√
6b2.

1.) By way of contradiction, let us assume that |p⃗| = |q⃗|. This yields

9a2 − 12ab+ 6b2 = c2 − 2bc+ b2 + 8a2

9a2 − 12ab+ 6b2 = a2 + b2 − 2bc+ b2 + 8a2

bc− 6ab+ 2b2 = 0.

Using the Euclid’s formula, we get

(2mn)(m2 + n2)− 6(m2 − n2)(2mn) + 2(2mn) = 0

m2 + n2 − 6m2 + 6n2 + 1 = 0

7n2 + 1 = 5m5

which implies that 7n2 + 1 ≡ 0 (mod 5) ⇐⇒ n2 ≡ 2 (mod 5). However, this congruence
has no solutions in N, hence |p⃗| ≠ |q⃗|.

2.) Further, we assume that |p⃗| = |r⃗|. Hence,

9a2 − 12ab+ 6b2 = 6b2

3a− 4b = 0

3a = 4b.

However, 3a is odd and 4b is even, a contradiction.
3.) Finally, we assume that |q⃗| = |r⃗|. Then

c2 − 2bc+ b2 + 8a2 = 6b2



REFERENCES 2140

c2 − 2bc− 5b2 + 8(c2 − b2) = 0

9c2 − 2bc− 12b2 = 0

9c2 = b(2c+ 13b)

where c is odd and b is even, which implies that the left side is odd, and the right side is
even, a contradiction.

Therefore, △P (R) is not an isosceles triangle.

Corollary 6. Let R be a primitive Pythagorean triple. The triangle △P (R) is not an
equilateral triangle.

5. Concluding remarks

To consider the three descendants of a primitive Pythagorean triple in a tree of prim-
itive Pythagorean triples as points in the three-dimensional space can provide further
insight into the structure of primitive Pythagorean triples. As these three descendants
form a triangle in either Berggren’s or Price’s tree, it is interesting to study the properties
of these triangles. This new approach offers numerous open problems, which we would
like to explore in our further research.

For example, we proved that the descendant triangle fails to be right triangle in both
Berggren’s and Price’s tree. If it always fails to be a right triangle, could it always be
acute? Is it obtuse in some cases? Can the set of all inner angles be somehow described?
These and further problems remain open and we intend to study them further.

Funding

This work was supported by European Union NextGenerationEUgrant through the
project vvgs-2023-3000.

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