J. Pure Appl. Algebra 216 (1) (2012) 184-191

Polynomial parametrization of Pythagorean quadruples, quintuples and sextuples.

Sophie Frisch (TU Graz) and Leonid Vaserstein (PSU)

Abstract. For n = 4 or 6, the Pythagorean n-tuples admit a parametrization by a single n-tuple of polynomials with integer coefficients (which is impossible for n = 3). For n = 5, there is an integer-valued polynomial Pythagorean 5-tuple which parametrizes Pythagorean 5-tuples (similar to the case n = 3). Pythagorean quadruples are closely related to (integer) Descartes quadruples, which we also parametrize by a Descartes quadruple of polynomials with integer coefficients.

Introduction

A Pythagorean triple is a triple of integers (x_{1}, x_{2}, x_{3}) satisfying x_{1}^{2} + x_{2}^{2} = x_{3}^{2}. More generally, for any integer n ≥ 3, and any commutative ring A, a Pythagorean n-tuple over A is an n-tuple (x_{1}, …, x_{n}) ∈ A^{n} such that

x_{1}^{2} + ⋯ + x_{n−1}^{2} = x_{n}^{2}.  (1)

Whenever A is not specified, we will understand A = ℤ. Likewise, “polynomial Pythagorean n-tuple” means a Pythagorean n-tuple over a ring of polynomials in finitely many indeterminates with coefficients in ℤ.

Instead of studying (1) directly, it is often convenient to substitute u = x_{n} + x_{n−1}, v = x_{n} − x_{n−1}, and to consider the equation

x_{1}^{2} + ⋯ + x_{n−2}^{2} = uv.  (2)

Over any ring A in which 2 is not a zero-divisor, this substitution and the reverse substitution

x_{n−1} = (u − v)/2,  x_{n} = (u + v)/2,  (3)

establish a bijection between solutions (x_{1}, …, x_{n}) ∈ A^{n} of (1) and solutions (x_{1}, …, x_{n−2}, u, v) ∈ A^{n} with u − v ∈ 2A of (2).

We recall the existing polynomial parametrizations of integer Pythagorean triples. It is well known that up to permutation of x_{1} and x_{2}, every Pythagorean triple has the form

(x_{1}, x_{2}, x_{3}) = y_{0}(2y_{1}y_{2}, y_{1}^{2} − y_{2}^{2}, y_{1}^{2} + y_{2}^{2})  (4)

with y_{i} ∈ ℤ. In other words, the set of integer Pythagorean triples is the union of f_{1}(ℤ^{3}) and f_{2}(ℤ^{3}) where

f_{1}(y_{0}, y_{1}, y_{2}) = y_{0}(2y_{1}y_{2}, y_{1}^{2} − y_{2}^{2}, y_{1}^{2} + y_{2}^{2})

and

f_{2}(y_{0}, y_{1}, y_{2}) = y_{0}(y_{1}^{2} − y_{2}^{2}, 2y_{1}y_{2}, y_{1}^{2} + y_{2}^{2})

are two Pythagorean triples over the polynomial ring ℤ[y_{0}, y_{1}, y_{2}]. We say that all Pythagorean triples are covered by two polynomial Pythagorean triples (in 3 parameters each).

It is easy to see that the intersection of f_{1}(ℤ^{3}) and f_{2}(ℤ^{3}) contains only the zero triple. We know that it is not possible to cover all Pythagorean triples by any one Pythagorean triple over ℤ[y_{1}, …, y_{m}] for any m [7]. It is, however, possible, to cover all integer Pythagorean triples by a single Pythagorean triple over the ring of integer-valued polynomials in 4 indeterminates [7]. An integer-valued polynomial is a polynomial with rational coefficients which takes integer values whenever the variables take integer values.

The primitive Pythagorean triples (x_{1}, x_{2}, x_{3}) with positive x_{3}, are, up to switching x_{1} and x_{2}, given by (4) with primitive (y_{1}, y_{2}) ∈ ℤ^{2} such that y_{1} + y_{2} is odd. The set of such pairs (y_{1}, y_{2}) admits a polynomial parametrization [15]. Thus, all primitive Pythagorean triples can be covered by 4 polynomial triples (in 95 parameters each, see [15], Example 14).

All positive Pythagorean triples are, up to switching of x_{1} and x_{2}, given by (4) with integers y_{1} > y_{2} > 0, y_{0} > 0. The set of such pairs admits a polynomial parametrization using the fact that every positive integer can be written as a sum of 4 squares plus 1. Thus, the positive Pythagorean triples can be covered by 2 polynomial Pythagorean triples in 12 parameters.

It is unknown whether the set of positive primitive Pythagorean triples can be parametrized by a finite set of polynomial Pythagorean triples.

1. Quadruples

After the short discussion of Pythagorean triples in the introduction, we now address the case n = 4, in other words, Pythagorean quadruples.

Pythagorean quadruples were described by Carmichael [1], Chpt. II, §10, as follows: up to permutation of x_{1}, x_{2}, x_{3}, every Pythagorean quadruple has the form

(x_{1}, x_{2}, x_{3}, x_{4}) = f(y_{0}, y_{1}, y_{2}, y_{3}, y_{4}) =
y_{0}(2y_{1}y_{3} + 2y_{2}y_{4}, 2y_{1}y_{4} − 2y_{2}y_{3}, y_{1}^{2} + y_{2}^{2} − y_{3}^{2} − y_{4}^{2}, y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2})  (5)

with integer values for the parameters y_{0}, …, y_{4}. Thus, all Pythagorean quadruples are covered by 6 polynomial Pythagorean quadruples (in 5 parameters). Considering the position of the odd entry, it is easy to see that at least 3 permutations of x_{1}, x_{2}, x_{3} are needed. If one examines Carmichael’s proof, one sees that three polynomial quadruples suffice, namely (x_{1}, x_{2}, x_{3}, x_{4}), (x_{1}, x_{3}, x_{2}, x_{4}) and (x_{3}, x_{2}, x_{1}, x_{4}).

We now show that a single polynomial Pythagorean quadruple covers all Pythagorean quadruples. Our proof does not make use of Carmichael’s result (but rather provides a shorter proof of Carmichael’s result as a byproduct). Nor do we use unique factorization in the ring of Gaussian integers ℤ[i] (which could be used to give an alternative proof).

Definition. An n-tuple w = (w_{1}, …, w_{n}) ∈ A^{n} is called unimodular if w_{1}A + ⋯ + w_{n}A = A. In the case when A = ℤ this means that gcd(w_{1}, …, w_{n}) = 1, i.e., w is primitive.

Proposition 1. The integer solutions of

x_{1}^{2} + x_{2}^{2} = uv  (6)

are parametrized by the polynomial quadruple

(x_{1}, x_{2}, u, v) = y_{0}(y_{1}y_{3} + y_{2}y_{4}, y_{1}y_{4} − y_{2}y_{3}, y_{1}^{2} + y_{2}^{2}, y_{3}^{2} + y_{4}^{2})  (7)

as the parameters vary through the integers. Also, y_{0} can be restricted to odd integers.

Proof. We represent the integer solutions of (6) as Hermitian matrices

w = ( \begin{matrix}u & x_{1} + ix_{2} \\ x_{1} − ix_{2} & v \end{matrix} ) = ( \begin{matrix}u & x \\ x̄ & v \end{matrix} )

of determinant 0 over the Gaussian integers ℤ[i].

The group GL(2, ℤ[i]) acts on the Hermitian matrices as follows

w → g^{∗}wg  (8)

where g ∈ GL(2, ℤ[i]) and * means transposition composed with entry-wise action of complex conjugation. In particular, for an elementary matrix g = E_{12}(λ) with λ = λ_{1} + λ_{2}i

( \begin{matrix}1 & 0 \\ λ̄ & 1 \end{matrix} ) ( \begin{matrix}u & x \\ x̄ & v \end{matrix} ) ( \begin{matrix}1 & λ \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}u & (x_{1} + λ_{1}u) + (x_{2} + λ_{2}u)i \\ (x_{1} + λ_{1}u) − (x_{2} + λ_{2}u)i & v + (λ_{1}^{2} + λ_{2}^{2})u + 2(λ_{1}x_{1} + λ_{2}x_{2}) \end{matrix} )

Setting either λ_{1} = 0 or λ_{2} = 0, we see that we can add an arbitrary integer multiple of u to x_{2}, leaving u and x_{1} unchanged, and we can add an arbitrary multiple of u to x_{1}, leaving u and x_{2} unchanged.

Given any solution with u ≠ 0, we can, by applying elementary matrices g in GL(2, ℤ[i]), make |x_{1}|, |x_{2}| ≤ |u|/2, and hence |v| ≤ |u| /2. Using the nontrivial permutation matrix in GL(2, ℤ[i]), we can switch u and v.

Therefore, by an argument of descent, the orbit under GL(2, ℤ[i]) of any unimodular solution w = (x_{1}, x_{2}, u, v) to (6) contains a solution with v = 0 (and hence x_{1} = x_{2} = 0) and u = 1 or −1. So we get

w = ( \begin{matrix}u & x_{1} + x_{2}i \\ x_{1} − x_{2}i & v \end{matrix} ) = g^{∗} ( \begin{matrix}c & 0 \\ 0 & 0 \end{matrix} ) g = ( ā ; b̄ ) c(a, b)  (9)

with c = ±1, where (a, b) is the first row of the matrix g ∈ GL(2, ℤ[i]).

So every integer solution of (6) has the form

w = ( \begin{matrix}u & x_{1} + x_{2}i \\ x_{1} − x_{2}i & v \end{matrix} ) = ( ā ; b̄ ) c(a, b)

with a, b ∈ ℤ[i], c ∈ ℤ. Conversely, every expression of this form is a solution to (6) – it is not necessary to restrict (a, b) to be primitive or c to be ±1 or (sum of 2 squares)-free.

Writing c = y_{0}, a = y_{1} + y_{2}i and b = y_{3} + y_{4}i, with indeterminates y_{k}, we obtain a polynomial solution

(u, v, x_{1}, x_{2}) = y_{0}(y_{1}^{2} + y_{2}^{2}, y_{3}^{2} + y_{4}^{2}, y_{1}y_{3} + y_{2}y_{4}, y_{1}y_{4} − y_{2}y_{3})

to (6) which covers all integer solutions. If we replace (a, b) above by (1 + i)(a, b), the solution is multiplied by 2. We can, therefore, restrict y_{0} to odd integers. □

Theorem 1. Let

f(y_{0}, y_{1}, y_{2}, y_{3}, y_{4}) = y_{0}(2y_{1}y_{3} + 2y_{2}y_{4}, 2y_{1}y_{4} − 2y_{2}y_{3}, y_{1}^{2} + y_{2}^{2} − y_{3}^{2} − y_{4}^{2}, y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2})

The polynomial Pythagorean quadruple

g = f(y_{0}/2, y_{1}, y_{2}, y_{3}, y_{1} + y_{2} + y_{3} + 2z) ∈ ℤ[y_{0}, y_{1}, y_{2}, y_{3}, z]

in 5 parameters covers all Pythagorean quadruples, i.e., the range of the function g: ℤ^{5} → ℤ^{4} consists of all Pythagorean quadruples.

Proof. Applying (3) to Proposition 1, we see that every Pythagorean quadruple has the form

(x_{1}, x_{2}, x_{3}, x_{4}) = y_{0}(y_{1}y_{3} + y_{2}y_{4}, y_{1}y_{4} − y_{2}y_{3}, (y_{1}^{2} + y_{2}^{2} − y_{3}^{2} − y_{4}^{2})/2, (y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2})/2)

where y_{i} ∈ ℤ and y_{0}(y_{1}+y_{2}+y_{3}+y_{4}) is even. Since y_{0} can be chosen odd, we may assume that y_{1}+y_{2}+y_{3}+y_{4} is even. Writing y_{4} = y_{1} + y_{2} + y_{3} + 2z, we prove Theorem 1. □

To get Carmichael’s result, note that x_{4} is odd for any primitive Pythagorean quadruple (x_{1}, x_{2}, x_{3}, x_{4}) and that exactly one of x_{1}, x_{2}, x_{3} is also odd. So we can make x_{3} + x_{4} even by switching, if necessary, x_{3} with x_{1} or x_{2}. Then gcd(x_{1}, x_{2}, u, v) = 2 for the corresponding solution (x_{1}, x_{2}, u, v) of (6). Going back from (7) with y_{0} = 2 to the Pythagorean quadruple, we obtain Carmichael’s formulas.

Notice that these formulas with y_{0} = 1 and primitive (y_{1}, y_{2}, y_{3}, y_{4}) do not necessarily give primitive solutions. Our proof shows that the necessary and sufficient condition for primitivity is the primitivity of (a, b) = (y_{1} + y_{2}i, y_{3} + y_{4}i). The set of primitive pairs of Gaussian integers admits a polynomial parametrization by methods of [15], but this is beyond the scope of the present paper.

2. Sextuples

We discuss Pythagorean sextuples before quintuples because we will use sextuples in the proof of the parametrization of Pythagorean quintuples by a single quintuple of integer-valued polynomials in the next section.

Dickson [4], Section 106, attempted to describe all Pythagorean sextuples, i.e., all integer solutions to

x_{1}^{2} + … + x_{5}^{2} = x_{6}^{2}  (10)

He observed that every integer solution of (10) gives rise to an integer solution of

x_{1}^{2} + … + x_{4}^{2} = uv,  (11)

He solves the equation (11), in a lengthy proof of some 6 pages, but then fails to address the question of the reverse substitution: how to return to Pythagorean sextuples from integer solutions of (11).

We will also start by parametrizing the integer solutions of (11), giving a short proof using quaternions.

Definition. The algebra of Lipschitz quaternions is the ℤ-algebra L generated by two symbols i and j subject to the defining relations i^{2} = −1, j^{2} = −1, and ji = −ij. We set k = ij.

We recall a few facts about the algebra of Lipschitz quaternions. L is a free ℤ-module with basis 1, i, j, k and a free ℤ[i]-module with basis 1, j. L can be represented as an algebra of 4 × 4 integer matrices or as an algebra of 2 × 2 matrices over ℤ[i] by identifying w = a + bi + cj + dk with

M_{4}(w) = ( \begin{matrix}a & b & c & d \\ −b & a & −d & c \\ −c & d & a & −b \\ −d & −c & b & a \end{matrix} ) or M_{2}(w) = ( \begin{matrix}a + bi & c + di \\ −c + di & a − bi \end{matrix} )

respectively.

An involution on L is given by the ℤ-algebra anti-isomorphism

a + bi + cj + dk ↦ (a + bi + cj + dk)^{∗} = a − bi − cj − dk.

In the 4×4 integer matrix representation this corresponds to transposition; and in the 2×2 Gaussian integer matrix representation, to transposition followed by complex conjugation.

Definition. The norm of w = a + bi + cj + dk ∈ L is defined as

(a^{2} + b^{2} + c^{2} + d^{2})^{2} = det(M_{4}(w)) = (det M_{2}(w))^{2}

and the reduced norm as

a^{2} + b^{2} + c^{2} + d^{2} = w^{∗}w = det M_{2}(w).

For a 2 × 2 matrix M over L we define the norm of M as the determinant of the 8 × 8 integer matrix obtained by replacing each matrix entry w by M_{4}(w), and the reduced norm as the determinant of the 4 × 4 matrix over ℤ[i] obtained by replacing each matrix entry w by M_{2}(w).

Remark. If w is a Hermitian 2 × 2 matrix over L, its entries commute and we can calculate the determinant in a naïve way, as

det ( \begin{matrix}u & x_{1} + x_{2}i + x_{3}j + x_{4}k \\ x_{1} − x_{2}i − x_{3}j − x_{4}k & v \end{matrix} ) = uv − x_{1}^{2} − x_{2}^{2} − x_{3}^{2} − x_{4}^{2}.

The reduced norm of w as defined above is the square of this determinant.

Proposition 2. A parametrization of all integer solutions (x_{1}, x_{2}, x_{3}, x_{4}, u, v) of

x_{1}^{2} + … + x_{4}^{2} = uv

in 9 parameters is given by

x_{1} = y_{0}(y_{1}y_{5} + y_{2}y_{6} + y_{3}y_{7} + y_{4}y_{8})
x_{2} = y_{0}(−y_{1}y_{6} + y_{2}y_{5} + y_{3}y_{8} − y_{4}y_{7})
x_{3} = y_{0}(−y_{1}y_{7} − y_{2}y_{8} + y_{3}y_{5} + y_{4}y_{6})
x_{4} = y_{0}(−y_{1}y_{8} + y_{2}y_{7} − y_{3}y_{6} + y_{4}y_{5})
u = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2})
v = y_{0}(y_{5}^{2} + y_{6}^{2} + y_{7}^{2} + y_{8}^{2}),

as the parameters y_{0}, …, y_{8} range through the integers. Here y_{0} may be restricted to ±1.

Proof. We identify integer solutions w = (x_{1}, x_{2}, x_{3}, x_{4}, u, v) of x_{1}^{2} + … + x_{4}^{2} = uv with 2 × 2 Hermitian matrices over the algebra of Lipschitz quaternions L

w = ( \begin{matrix}u & x_{1} + x_{2}i + x_{3}j + x_{4}k \\ x_{1} − x_{2}i − x_{3}j − x_{4}k & v \end{matrix} )

of reduced norm 0.

The group GL(2, L) acts on the set of 2×2 Hermitian matrices over L of reduced norm 0 by (g, w) ↦ g^{∗}wg, for g ∈ GL(2, L), where g^{∗} results from g by application of the involution ∗ to each entry, followed by transposition.

Given any unimodular solution of (11) with u ≠ 0, using an elementary matrix in GL(2, L), we can make |x_{1}|, |x_{2}|, |x_{3}|, |x_{4}| ≤ |u|/2, and hence |v| ≤ |u|. The inequality is strict unless |x_{m}| = |v|/2 for m = 1, 2, 3, 4, in which case |v| = 2 by unimodularity. In this last case, using an elementary matrix, we can arrange x_{m} = 1 for m = 1, 2, 3, 4.

Using the nontrivial permutation matrix in GL(2, L), we can switch u and v. Therefore, by induction on |u|, the orbit of any unimodular solution w = (x_{1}, x_{2}, x_{3}, x_{4}, u, v) to (11) contains a solution with either |u| = 1 and x_{m} = 0 for m = 1, 2, 3, 4 or |u| = 2 and x_{m} = 1 for m = 1, 2, 3, 4.

So we get that either

w = g^{∗} ( \begin{matrix}c & 0 \\ 0 & 0 \end{matrix} ) g = ( \begin{matrix}a^{∗} \\ b^{∗} \end{matrix} ) c(a, b)

with c = ±1 where (a, b) is the first row of the matrix g ∈ GL(2, L) or

w = ±g^{∗} ( \begin{matrix}2 & 1 + i + j + k \\ 1 − i − j − k & 2 \end{matrix} ) g = ± ( \begin{matrix}a^{∗} \\ b^{∗} \end{matrix} ) (a, b)

where (a, b) = ±(1 − i, 1 + j)g with g ∈ GL(2, L), because

( \begin{matrix}2 & 1 + i + j + k \\ 1 − i − j − k & 2 \end{matrix} ) = (1 − i, 1 + j)^{∗}(1 − i, 1 + j).

So every integer solution of (11) has the form

w = ( \begin{matrix}u & x_{1} + x_{2}i + x_{3}j + x_{4}k \\ x_{1} − x_{2}i − x_{3}j − x_{4}k & v \end{matrix} ) = ( \begin{matrix}a^{∗} \\ b^{∗} \end{matrix} ) c(a, b)

with a, b ∈ L and c ∈ ℤ. Here we need not restrict (a, b) to be primitive.

Writing c = y_{0}, a = y_{1} +y_{2}i+y_{3}j +y_{4}k and b = y_{5} +y_{6}i+y_{7}j +y_{8}k we obtain the desired parametrization of all solutions w = (x_{1}, x_{2}, x_{3}, x_{4}, u, v) of (11).

If we replace (a, b) above by d(a, b) with d ∈ L, the solution w is multiplied by d^{∗}d, which is equivalent to replacing y_{0} by y_{0}d^{∗}d. Since every nonnegative integer is of the form d^{∗}d (sum of 4 squares) we can restrict y_{0} to be ±1. □

Returning to Pythagorean n-tuples, the following polynomial Pythagorean sextuple is known:

(x_{1}, …, x_{6}) = h(y_{0}, …, y_{8}) ∈ ℤ[y_{0}, …, y_{8}]^{6}  (12)

with

x_{1} = 2y_{0}(y_{1}y_{5} + y_{2}y_{6} + y_{3}y_{7} + y_{4}y_{8})
x_{2} = 2y_{0}(−y_{1}y_{6} + y_{2}y_{5} + y_{3}y_{8} − y_{4}y_{7})
x_{3} = 2y_{0}(−y_{1}y_{7} − y_{2}y_{8} + y_{3}y_{5} + y_{4}y_{6})
x_{4} = 2y_{0}(−y_{1}y_{8} + y_{2}y_{7} − y_{3}y_{6} + y_{4}y_{4})
x_{5} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} − y_{5}^{2} − y_{6}^{2} − y_{7}^{2} − y_{8}^{2})
x_{6} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} + y_{5}^{2} + y_{6}^{2} + y_{7}^{2} + y_{8}^{2})

There are, however, integer Pythagorean sextuples such as (1, 1, 1, 1, 0, 2) that do not arise from the above polynomial sextuple with integer parameters y_{i}. We now give a parametrization of all integer Pythagorean sextuples by a single polynomial Pythagorean sextuple in 9 parameters, or, by restricting the parameter y_{0} to ±1, a parametrization by two integer Pythagorean sextuples in 8 parameters each.

Theorem 2. Let h = h(y_{0}, …, y_{8}) ∈ ℤ[y_{0}, …, y_{8}]^{6} be the polynomial Pythagorean sextuple (12) above. Then the polynomial Pythagorean sextuple in ℤ[y_{0}, …, y_{7}, z]^{6}

g(y_{0}, y_{1}, y_{2}, y_{3}, y_{4}, y_{5}, y_{6}, y_{7}, z) = h(y_{0}/2, y_{1}, y_{2}, y_{3}, y_{4}, y_{5}, y_{6}, y_{7}, y_{1} + y_{2} + y_{3} + y_{4} + y_{5} + y_{6} + y_{7} + 2z)

in 9 parameters covers all Pythagorean sextuples, i.e., the range of the function g: ℤ^{9} → ℤ^{6} is precisely the set of all Pythagorean sextuples. Also, the parameter y_{0} can be restricted to ±1.

Proof. We obtain all Pythagorean sextuples from all solutions of (11) with u−v even by by (3). Since we can take y_{0} odd (even ±1) in Proposition 2, we may assume that y_{1}+⋯+y_{8} is even. Writing y_{8} = y_{1}+⋯+y_{7}+2z, we obtain a Pythagorean sextuple over ℤ[y_{0}, …, y_{7}, z] in 9 parameters which parametrizes all Pythagorean sextuples:

x_{1} = y_{0}(y_{1}y_{5} + y_{2}y_{6} + y_{3}y_{7} + y_{4}(y_{1} + ⋯ + y_{7} + 2z))  (13)
x_{2} = y_{0}(−y_{1}y_{6} + y_{2}y_{5} + y_{3}(y_{1} + ⋯ + y_{7} + 2z) − y_{4}y_{7})
x_{3} = y_{0}(−y_{1}y_{7} − y_{2}(y_{1} + ⋯ + y_{7} + 2z) + y_{3}y_{5} + y_{4}y_{6})
x_{4} = y_{0}(−y_{1}(y_{1} + ⋯ + y_{7} + 2z) + y_{2}y_{7} − y_{3}y_{6} + y_{4}y_{5})
x_{5} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} − y_{5}^{2} − y_{6}^{2} − y_{7}^{2} − (y_{1} + ⋯ + y_{7} + 2z)^{2})/2
x_{6} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} + y_{5}^{2} + y_{6}^{2} + y_{7}^{2} + (y_{1} + ⋯ + y_{7} + 2z)^{2})/2

□

Remark. The ℤ-algebra L is a subring of the ring H of rational quaternions (or Hamilton quaternions). Adjoining to L the element (1 + i + j + k)/2, we obtain a larger subring L′ of H, called the ring of Hurwitz quaternions. This ring L′ has certain unique factorization properties (cf. [2]), which, however, we did not use. They could be used to give alternative proofs for the results in this section.

3. Quintuples

We now consider the case n = 5 of Pythagorean quintuples. We obtain Theorem 3 from Proposition 2 via Proposition 3.

Proposition 3. A parametrization of all integer quintuples (x_{1}, x_{2}, x_{3}, u, v) satisfying

x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = uv

by a quintuple of polynomials with integer coefficients in the 12 parameters y_{0}, z_{0}, z_{1}, z_{2}, z_{3}, z_{4}, z_{12}, z_{13}, z_{14}, z_{23}, z_{24}, z_{34} is given by

x_{1} = y_{0}(y_{1}y_{5} + y_{2}y_{6} + y_{3}y_{7} + y_{4}y_{8})
x_{2} = y_{0}(−y_{1}y_{6} + y_{2}y_{5} + y_{3}y_{8} − y_{4}y_{7})
x_{3} = y_{0}(−y_{1}y_{7} − y_{2}y_{8} + y_{3}y_{5} + y_{4}y_{6})
u = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2})
v = y_{0}(y_{5}^{2} + y_{6}^{2} + y_{7}^{2} + y_{8}^{2}),

with

y_{1} = z_{0}z_{1}, y_{2} = z_{0}z_{2}, y_{3} = z_{0}z_{3}, y_{4} = z_{0}z_{4},
y_{5} = −z_{14}z_{1} − z_{24}z_{2} − z_{34}z_{3}
y_{6} = z_{13}z_{1} + z_{23}z_{2} − z_{34}z_{4}
y_{7} = −z_{12}z_{1} + z_{23}z_{3} + z_{24}z_{4}
y_{8} = −z_{12}z_{2} − z_{13}z_{3} − z_{14}z_{4}.

Proof. To solve x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = uv, we set x_{4} = 0 in the general solution to x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = uv obtained in Proposition 2:

−y_{1}y_{8} + y_{2}y_{7} − y_{3}y_{6} + y_{4}y_{5} = 0  (14)

(The case y_{0} = 0 only contributes the zero solution which we will not miss.)

The integer solutions of (14) can be parametrized by 11 parameters as follows.

First we write (y_{1}, y_{2}, y_{3}, y_{4}) = z_{0}(z_{1}, z_{2}, z_{3}, z_{4}) with z_{i} ∈ ℤ and (z_{1}, z_{2}, z_{3}, z_{4}) unimodular such that

(z_{1}, z_{2}, z_{3}, z_{4}).(−y_{8}, y_{7}, −y_{6}, y_{5}) = 0

(the case y_{1} = y_{2} = y_{3} = y_{4} = 0 only contributes solutions (0, 0, 0, 0, v) which we will not miss). By [14], Remark after Lemma 9.6, we can write

(−y_{8}, y_{7}, −y_{6}, y_{5}) = z_{12}(z_{2}, −z_{1}, 0, 0)+z_{13}(z_{3}, 0, −z_{1}, 0)+z_{14}(z_{4}, 0, 0, −z_{1})+z_{23}(0, z_{3}, −z_{2}, 0)+z_{24}(0, z_{4}, 0, −z_{2})+z_{34}(0, 0, z_{4}, −z_{3}).

This gives a parametrization of the integer solutions of (14) in the 11 parameters z_{0}, z_{1}, z_{2}, z_{3}, z_{4}, z_{12}, z_{13}, z_{14}, z_{23}, z_{24}, z_{34}.

Therefore all integer solutions of x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = uv are parametrized by a polynomial solution with 12 parameters including y_{0}. □

Another parametrization of x_{1}^{2} +x_{2}^{2} +x_{3}^{2} = uv with 20 parameters can be obtained using [15], Proposition 3.4 with k = 8. We now parametrize integer Pythagorean quintuples by a single Pythagorean quintuple over the ring of integer-valued polynomials in 14 variables. This can be used to construct a parametrization by a finite number of integer-coefficient polynomial Pythagorean quintuples [6]. Whether it is possible to parametrize integer Pythagorean quintuples by a single quintuple of integer-coefficient polynomials or not, we do not know.

Theorem 3. A parametrization of all Pythagorean quintuples by a quintuple of integer-valued polynomials in the 14 variables w_{0}, w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34}, t_{1}, t_{2}, t_{3}, d_{1}, d_{2}, d_{3}, w_{4} is given by (f_{1}, f_{2}, f_{3}, f_{5}, f_{6}), where

f_{1} = 2y_{0}(y_{1}y_{5} + y_{2}y_{6} + y_{3}y_{7} + y_{4}y_{8})
f_{2} = 2y_{0}(−y_{1}y_{6} + y_{2}y_{5} + y_{3}y_{8} − y_{4}y_{7})
f_{3} = 2y_{0}(−y_{1}y_{7} − y_{2}y_{8} + y_{3}y_{5} + y_{4}y_{6})
f_{5} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} − y_{5}^{2} − y_{6}^{2} − y_{7}^{2} − y_{8}^{2})/2
f_{6} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} + y_{5}^{2} + y_{6}^{2} + y_{7}^{2} + y_{8}^{2})/2

and

y_{1} = z_{0}z_{1}, y_{2} = z_{0}z_{2}, y_{3} = z_{0}z_{3}, y_{4} = z_{0}z_{4},
y_{5} = −z_{14}z_{1} − z_{24}z_{2} − z_{34}z_{3}
y_{6} = z_{13}z_{1} + z_{23}z_{2} − z_{34}z_{4}
y_{7} = −z_{12}z_{1} + z_{23}z_{3} + z_{24}z_{4}
y_{8} = −z_{12}z_{2} − z_{13}z_{3} − z_{14}z_{4}.

and further

z_{0} =w_{0} + t_{1}w_{0} + t_{2}w_{0} − 2t_{1}t_{2}w_{0} + t_{3}w_{0} − 2t_{1}t_{3}w_{0} − t_{2}t_{3}w_{0} + 2t_{1}t_{2}t_{3}w_{0} + t_{1}w_{12} − t_{1}t_{2}w_{12}−
− t_{1}t_{3}w_{12} + t_{2}t_{3}w_{12} + t_{2}w_{13} − t_{1}t_{2}w_{13} + t_{3}w_{14} − t_{1}t_{3}w_{14} + t_{1}w_{23} + t_{2}w_{23} − 2t_{1}t_{2}w_{23}−
− t_{1}t_{3}w_{23} − t_{2}t_{3}w_{23} + 2t_{1}t_{2}t_{3}w_{23} + t_{1}w_{24} − t_{1}t_{2}w_{24} + t_{3}w_{24} − 2t_{1}t_{3}w_{24} − t_{2}t_{3}w_{24}+
+ 2t_{1}t_{2}t_{3}w_{24} + t_{2}w_{34} − t_{1}t_{2}w_{34} + t_{3}w_{34} − t_{1}t_{3}w_{34} − 2t_{2}t_{3}w_{34} + 2t_{1}t_{2}t_{3}w_{34}
z_{1} =2d_{1} + t_{1}t_{2} + t_{3} − 2t_{1}t_{2}t_{3} + w_{4}
z_{2} =2d_{2} + t_{1} − t_{1}t_{2} + t_{3} − t_{1}t_{3} − t_{2}t_{3} + 2t_{1}t_{2}t_{3} + w_{4}
z_{3} =2d_{3} + t_{2} + t_{3} − t_{1}t_{3} − 2t_{2}t_{3} + 2t_{1}t_{2}t_{3} + w_{4}
z_{4} =w_{4}
z_{12} =w_{12} + t_{1}t_{2}w_{12} − t_{1}t_{2}t_{3}w_{12} + t_{1}t_{2}w_{14} − t_{1}t_{2}t_{3}w_{14} + t_{1}t_{2}w_{23} − t_{1}t_{2}t_{3}w_{23} + t_{1}t_{2}w_{34} − t_{1}t_{2}t_{3}w_{34}
z_{13} =w_{13} + t_{1}t_{3}w_{13} − t_{1}t_{2}t_{3}w_{13} + t_{1}t_{3}w_{14} − t_{1}t_{2}t_{3}w_{14} + t_{1}t_{3}w_{23} − t_{1}t_{2}t_{3}w_{23} + t_{1}t_{3}w_{24} − t_{1}t_{2}t_{3}w_{24}
z_{14} =w_{14}
z_{23} =w_{23}
z_{24} =t_{1}t_{2}t_{3}w_{12} + t_{1}t_{2}t_{3}w_{13} + w_{24} + t_{1}t_{2}t_{3}w_{24} + t_{1}t_{2}t_{3}w_{34}
z_{34} =w_{34}

Proof. To go from the solutions of x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = uv parametrized in Proposition 3 to the solutions of x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = x_{5}^{2} we use (3), allowing only those u, v with u ± v even.

In our case, we need to parametrize those z_{0}, …, z_{34} that make y_{1}+y_{2}+…+y_{8} even, i.e., those z_{0}, …, z_{34} such that E = z_{0}z_{1} + z_{0}z_{2} + z_{0}z_{3} + z_{0}z_{4} + −z_{14}z_{1} − z_{24}z_{2} − z_{34}z_{3} − z_{13}z_{1} − z_{23}z_{2} + z_{34}z_{4} − z_{12}z_{1} + z_{23}z_{3} + z_{24}z_{4} − z_{12}z_{2} − z_{13}z_{3} − z_{14}z_{4} is even.

This is achieved by the following substitution, which, after simplification, gives the parametrization in the statement of the theorem.

(z_{0}, z_{1}, z_{2}, z_{3}, z_{4}, z_{12}, z_{13}, z_{14}, z_{23}, z_{24}, z_{34}) =
(w_{0}, w_{4} + 2d_{1}, w_{4} + 2d_{2}, w_{4} + 2d_{3}, w_{4}, w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34})(1 − t_{1})(1 − t_{2})(1 − t_{3})+

(w_{14}+w_{24}+w_{34}+2w_{0}, w_{4}+2d_{1}+1, w_{4}+2d_{2}+1, w_{4}+2d_{3}+1, w_{4}, w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34})(1−t_{1})(1−t_{2})t_{3}+ (w_{13} + w_{23} + w_{34} + 2w_{0}, w_{4} + 2d_{1}, w_{4} + 2d_{2}, w_{4} + 2d_{3} + 1, w_{4}, w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34})(1 − t_{1})t_{2}(1 − t_{3})+ (w_{12} + w_{23} + w_{24} + 2w_{0}, w_{4} + 2d_{1}, w_{4} + 2d_{2} + 1, w_{4} + 2d_{3}, w_{4}, w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34})t_{1}(1 − t_{2})(1 − t_{3})+

(w_{12} + w_{13} + w_{14} + 2w_{0}, w_{4} + 2d_{1} + 1, w_{4} + 2d_{2}, w_{4} + 2d_{3}, w_{4}, w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34})(1 − t_{1})t_{2}t_{3}+ (w_{0}, w_{4} + 2d_{1} + 1, w_{4} + 2d_{2} + 1, w_{4} + 2d_{3}, w_{4}, w_{12}, w_{23} + w_{24} + w_{14} + 2w_{13}, w_{14}, w_{23}, w_{24}, w_{34})t_{1}(1 − t_{2})t_{3}+ (w_{0}, w_{4} + 2d_{1} + 1, w_{4} + 2d_{2}, w_{4} + 2d_{3} + 1, w_{4}, w_{23} + w_{14} + w_{34} + 2w_{12}, w_{13}, w_{14}, w_{23}, w_{24}, w_{34})t_{1}t_{2}(1 − t_{3})+

(w_{0}, w_{4} + 2d_{1}, w_{4} + 2d_{2} + 1, w_{4} + 2d_{3} + 1, w_{4}, w_{12}, w_{13}, w_{14}, w_{23}, w_{12} + w_{13} + w_{34} + 2w_{24}, w_{34})t_{1}t_{2}t_{3}

□

4. Descartes quadruples

In 1643 Descartes [3] described a relationship between the radii of four mutually tangent circles (called a Descartes configuration), namely,

2(b_{1}^{2} + b_{2}^{2} + b_{3}^{2} + b_{4}^{2}) = (b_{1} + b_{2} + b_{3} + b_{4})^{2}  (15)

where b_{i} are the reciprocals of the radii. Others, including Steiner, Beecroft, and Soddy [13], rediscovered the result. We call an integer solution of (15) a Descartes quadruple.

Given one Descartes configuration, there is a geometric way to produce plenty of them creating an Apollonian packing. If the four initial curvatures b_{i} are integers, all curvatures in the packing are integers. There are several publications about integer Apollonian packings [8], [9], [10], [5],[12], [11]. A bijection between integer Pythagorean quadruples and integer Descartes quadruples can be found in [8], Lemma 2.1.

In this section we parametrize all integer solutions of (15) by a single polynomial solution in 5 parameters, using a bijection between integer Descartes quadruples and integer solutions of (6).

Given an integer solution (x_{1}, x_{2}, u, v) of (6),

b_{1} = u + v − 2x_{1} + x_{2}, b_{2} = u + x_{2}, b_{3} = v + x_{2}, b_{4} = −x_{2}  (16)

is an integer solution of (15). Conversely, we can invert this linear transformation: given an integer solution (b_{1}, b_{2}, b_{3}, b_{4}) of (15), b_{1} + b_{2} + b_{3} + b_{4} is even and

x_{1} = (−b_{1} + b_{2} + b_{3} + b_{4})/2, x_{2} = −b_{4}, u = b_{2} + b_{4}, v = b_{3} + b_{4}

is an integer solution of (6).

Theorem 4. A parametrization of all integer solutions of

2(b_{1}^{2} + b_{2}^{2} + b_{3}^{2} + b_{4}^{2}) = (b_{1} + b_{2} + b_{3} + b_{4})^{2}  (15)

in 5 parameters is given by

b_{1} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} − 2y_{1}y_{3} − 2y_{2}y_{4} + y_{1}y_{4} − y_{2}y_{3}),
b_{2} = y_{0}(y_{1}^{2} + y_{2}^{2} + y_{1}y_{4} − y_{2}y_{3}),
b_{3} = y_{0}(y_{3}^{2} + y_{4}^{2} + y_{1}y_{4} − y_{2}y_{3}),
b_{4} = y_{0}(−y_{1}y_{4} − y_{2}y_{3}).

Proof. In the expression (16) of b_{1}, b_{2}, b_{3}, b_{4} in terms of a solution x_{1}, x_{2}, u, v of (6) we have substituted the parametrization of all integer solutions of (6) from Proposition 1. □

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