ALGEBRAIC STRUCTURES OF SOME SETS OF PYTHAGOREAN TRIPLES II

Marek Wójtowicz

Abstract. A natural bijection from ℤ^{2} onto the set of all Pythagorean triples P = {(a, b, c) ∈ ℤ^{3} : a^{2} +b^{2} = c^{2}} is given (Theorem 6). Consequently, all algebraic structures of ℤ^{2} are carried in a natural way onto P (Theorems 7, 8, and 9). This solves the open problem of defining ring operations under which P is essentially a different ring than the one constructed by B. Dawson (Example, Section 5). This article and the enumeration of its sections and theorems is a continuation of the author’s paper [3].

4. Elements of P. Corollary 1 given in Section 2 yields the useful base for a description of all sets P_{n}, n ≠ 0 (the case n = 0 is obvious), and their elements by giving the apparent form of natural numbers x_{n} ≥ 2 with ψ_{n}(P_{n}) = x_{n}ℤ (Proposition 1 below). To this end we define two functions ℤ → ℤ: quasi– square root ^{∗} √ , and degree of evenness d_{ev}, respectively, as follows.

If p_{1}^{α_{1}} · . . . · p_{k}^{α_{k}} is the prime factorization of |n| ≥ 1, then \sqrt[∗]{n} := p_{1}^{β_{1}} · . . . · p_{k}^{β_{k}}, where β_{j} = [(α_{j} + 1)/2], j = 1, …, k, and [x] denotes the integer part of x, and \sqrt[∗]{0} := 0.

If n = 2^{r}n_{0}, where r, n_{0} ∈ ℤ, r ≥ 0 and (2, n_{0}) = 1, then

d_{ev}(n) = \{ \begin{matrix}1, & for r odd, \\ 2, & for r even and r ≥ 0. \end{matrix}

(A function similar to quasi–square root was used by Dawson in [1].) In the proofs of Propositions 1 and 2 we shall use the following properties of ^{∗} √ and d_{ev} (recall that if p_{1}^{α_{1}} · . . . · p_{k}^{α_{k}} is the prime factorization of n ∈ ℕ, then the square–free kernel of n is defined as t(n) = p_{1} · . . . · p_{k}, and n is called a square–free number provided that t(n) = n).

(1) 0 ≤ (\sqrt[∗]{n})^{2} ≤ |n|t(|n|).

(2) If n | a^{2}, for some a ∈ ℤ, then \sqrt[∗]{n} | a.

(3) For every integer n, l, p with l ≥ 0 and p prime we have \sqrt[∗]{n^{2}} = |n| and \sqrt[∗]{p^{2l+1}} = p^{l+1}.

(4) n/\sqrt[∗]{n} and (\sqrt[∗]{n})^{2}/n are integers for all n ≠ 0.

(5) We have \sqrt[∗]{nm} ≤ \sqrt[∗]{n}\sqrt[∗]{m} for all n, m ∈ ℤ, and \sqrt[∗]{nm} = \sqrt[∗]{n}\sqrt[∗]{m} whenever (n, m) = 1.

(6) If n ≠ 0, then (\sqrt[∗]{n})^{2}/n is even if and only if d_{ev}(n) = 1 (and so (\sqrt[∗]{n})^{2}/n is odd if and only if d_{ev}(n) = 2).

(7) For all s ∈ ℤ we have d_{ev}(2s + 1) = 2, and d_{ev}(4s + 2) = 1.

(8) \sqrt[∗]{n} = |n| if and only if |n| is square–free.

(9) If n, r ∈ ℕ with r odd and (n, r) = 1, then d_{ev}(nr) = d_{ev}(n).

\underline{Proposition 1}. Let ψ_{n} be the ring isomorphism defined in Theorem 1 and acting from P_{n} onto the ring ideal G_{n} = x_{n}ℤ, where x_{n} is some integer greater than or equal to 2 and n ≠ 0. Then x_{n} = d_{ev}(n)\sqrt[∗]{n}. In particular, x_{n} = x_{2n} = 2\sqrt[∗]{n} for n odd, and hence, x_{r} = x_{2r} = 2r for every odd and square–free number r ∈ ℕ.

\underline{Proof}. We shall consider the case n ≥ 1 only; for n ≤ −1 the proof is similar. We have ψ_{n}(P_{n}) = {a_{k}^{(n)} − n : a_{k}^{(n)} = kx_{n} + n and (a_{k}^{(n)}, b_{k}^{(n)}, c_{k}^{(n)}) ∈ P_{n}, k ∈ ℤ}, where b_{k}^{(n)} and c_{k}^{(n)} are defined in Section 1. Hence,

(10) a_{1}^{(n)} = x_{n} + n. Since (a_{1}^{(n)})^{2} − n^{2})/2n = b_{1}^{(n)} ∈ ℕ, we have n | (a_{1}^{(n)})^{2}; hence, by property (2),

(11) a_{1}^{(n)} = t\sqrt[∗]{n} for some t ∈ ℕ. By (4), (10), and (11), we obtain

(12) x_{n} = l\sqrt[∗]{n}, where l = t − n/\sqrt[∗]{n} ∈ ℕ, and so, by (10), a_{1}^{(n)} = l\sqrt[∗]{n} + n; putting this value into the formula defining b_{1}^{(n)} we obtain

b_{1}^{(n)}(l) = ((\sqrt[∗]{n})^{2}/n) · (l/2) + l\sqrt[∗]{n}.

It follows that l equals the least natural number with b_{1}^{(n)}(l) ∈ ℕ. By properties (4) and (6) we have that l = 1 = d_{ev}(n) provided that (\sqrt[∗]{n})^{2}/n is even, and l = 2 = d_{ev}(n) for (\sqrt[∗]{n})^{2}/n odd. Finally, by (12), we get x_{n} = d_{ev}(n)\sqrt[∗]{n}. The particular case follows from properties (5), (7) and (8).

The main result of this section reads as follows.

\underline{Theorem 6}. For every (a, b, c) ∈ P^{∗} := P\P_{0} there exists exactly one pair (k, n) ∈ ℤ × (ℤ\{0}) with

a = a(k, n) = k · d_{ev}(n) · \sqrt[∗]{n} + n,

b = b(k, n) = k^{2} · \frac{(d_{ev}(n) · \sqrt[∗]{n})^{2}}{2n} + k · d_{ev}(n) · \sqrt[∗]{n},

c = c(k, n) = b(k, n) + n.

Conversely, for every (k, n) ∈ ℤ × (ℤ\{0}), the triple (a, b, c), where the numbers a, b, c are defined as above, is an element of P^{∗}. Consequently, the function α: ℤ^{2} → P, given by the rule

α(k, n) = \{ \begin{matrix}(a(k, n), b(k, n), c(k, n)) & for n ≠ 0 \\ (0, k, k) & for n = 0 \end{matrix}

maps both ℤ^{2} onto P and ℤ × (ℤ\{0}) onto P^{∗} bijectively. The inverse function α^{−1}: P → ℤ^{2} is of the form: (a, b, c) → (k, n), where

k = k(a, b, c) = \frac{a + b − c}{d_{ev}(c − b) · \sqrt[∗]{c − b}} and n = n(a, b, c) = c − b for c − b ≠ 0,

and k(0, j, j) = j and n(0, j, j) = 0.

\underline{Proof}. Since P = ∪_{n∈ℤ}P_{n}, and since the sets P_{n}, n ∈ ℤ, are pairwise disjoint, by Proposition 1 and the presence of elements of P given in Section 1, the first part of the theorem is clear. To prove the second part we must show that b(k, n) and c(k, n) are integers for all k, n ∈ ℤ. Assume that n ≠ 0 (the case n = 0 is trivial), and observe that the number d_{ev}(n)\sqrt[∗]{n} is even, whence, by property (4), s(n) := (d_{ev}(n)\sqrt[∗]{n})^{2}/2n is an integer. It follows that for every k ∈ ℤ we have b(k, n) = k^{2}s(n) + kd_{ev}(n)\sqrt[∗]{n} ∈ ℤ, and hence, c(k, n) = b(k, n) + n ∈ ℤ also. If (a, b, c) ∈ P^{∗}, then c − b = n ≠ 0, hence, by the formula defining a(k, n), we have k = k(a, b, c) = (a + b − c)/(d_{ev}(c − b) · \sqrt[∗]{c − b}); hence, we get the form of α^{−1}.

5. The Ring and the Lattice Structures on P . Now we are in position to transfer two ring structures from ℤ^{2} onto P: the “coordinatewise” one, where the additive zero is (0, 0) and the multiplicative unit is (1, 1), and the complex one, with the additive zero defined as above and the multiplicative unit equals (1, 0). The reader should note that the ring operations presented below are different from those constructed by B. Dawson (see Example below), and this solves affirmatively the open problem stated in [1], to define other operations in a natural way (cf. the remark after Lemma in Section 1) under which P is essentially a different ring than Dawson’s. Moreover, the operations given in Theorem 7 do not extend Grytczuk’s operations (i.e., when restricted to P_{n}, n ≠ 0, these operations take values outside of P_{n}, in general, and the present author could not find any satisfactory extensions of Grytczuk’s operations). Nevertheless, Dawson’s multiplicative unit of P and the unit given in Theorem 7 (i) are identical; we denote this particular triple (3, 4, 5) as 1_{P}.

The two theorems given below are now immediate consequences of Theorem 6 and the Lemma.

\underline{Theorem 7}. The set P is a commutative ring with unit under the following pairs of addition and multiplication:

(i) (a_{1}, b_{1}, c_{1}) ⊕ (a_{1}, b_{1}, c_{1}) := α(k_{1} + k_{2}, n_{1} + n_{2}), and (a_{1}, b_{1}, c_{1}) ⊙ (a_{1}, b_{1}, c_{1}) := α(k_{1}k_{2}, n_{1}n_{2}), with the additive zero (0, 0, 0) = α(0, 0) and the multiplicative unit 1_{P} = α(1, 1);

(ii) (a_{1}, b_{1}, c_{1}) ⊕ (a_{1}, b_{1}, c_{1}) := α(k_{1} + k_{2}, n_{1} + n_{2}), and (a_{1}, b_{1}, c_{1}) ⊙ (a_{1}, b_{1}, c_{1}) := α(k_{1}k_{2} − n_{1}n_{2}, k_{1}n_{2} + k_{2}n_{1}), with the additive zero (0, 0, 0) and the multiplicative unit (0, 1, 1) = α(1, 0), where the numbers k_{j} = k(a_{j}, b_{j}, c_{j}) and n_{j} = n(a_{j}, b_{j}, c_{j}), j = 1, 2, and the function α are defined in Theorem 6.

\underline{Theorem 8}. The set P is a distributive lattice under the following partial ordering:

(a_{1}, b_{1}, c_{1}) ≤ (a_{2}, b_{2}, c_{2}) if and only if
k(a_{1}, b_{1}, c_{1}) ≤ k(a_{2}, b_{2}, c_{2}) and n(a_{1}, b_{1}, c_{1}) ≤ n(a_{1}, b_{1}, c_{1}),

where the numbers k(a_{j}, b_{j}, c_{j}) and n(a_{j}, b_{j}, c_{j}), for j = 1, 2, are defined in Theorem 6.

\underline{Example}. We shall show that Dawson’s operations denoted here by ⊕_{D} and ⊙_{D}, differ from those given in Theorem 7. In case (i), we have 1_{P} ⊕ 1_{P} = α(1, 1) ⊕ α(1, 1) = α(2, 2) = (6, 8, 10) = 2 · 1_{P}, and (4, 3, 5) ⊙ (4, 3, 5) = α(1, 2) ⊙ α(1, 2) = α(1, 4) = (8, 6, 10) = 2 · (4, 3, 5). On the other hand, from [1] it follows that 1_{P} ⊕_{D} 1_{P} = (4, 3, 5), and that (4, 3, 5) ⊙_{D} (4, 3, 5) = (16, 30, 32) = 2 · (8, 15, 16).

In case (ii), the addition is identical as in case (i), and we have (4, 3, 5) ⊙ (4, 3, 5) = α(1, 2) ⊙ α(1, 2) = α(−3, 4) = (−8, 6, 10) = 2 · (−4, 3, 5).

6. The Field Structure on P^{∗}. The following observation leads to the construction of the field structure on some classes of subsets of P. Theorem 6 and properties (7) and (8) yield

\underline{Proposition 2}. For every odd and square–free number r ≥ 1 we have α(1, r) = r1_{P}. In particular, for each pair of different odd prime numbers p, q we have α(1, pq) = pq1_{P}.

This result suggests to find a multiplication ∘ on ℤ × (ℤ\{0}) with (1, p) ∘ (1, q) = (1, pq) for all odd prime numbers p, q and to carry it onto P^{∗}, with the help of the function α, to obtain the equation α((1, p) ∘ (1, q)) = α(1, p) ⊙ α(1, q). There exist two multiplications which fulfill that requirement: the coordinatewise multiplication, just used in Theorem 7 (i), and the multiplication connected with fractions, where the pair (1, n) corresponds to the fraction 1/n. Let R denote the equivalence relation on ℤ × (ℤ\{0}) of the form

(k_{1}, n_{1})R(k_{2}, n_{2}) if and only if k_{1}n_{2} = k_{2}n_{1},

and let R(α) be carried from ℤ × (ℤ\{0}) onto P^{∗}, by means of α, relation R, i.e.

(a_{1}, b_{1}, c_{1})R(α)(a_{2}, b_{2}, c_{2}) if and only if
(a_{1} + b_{1} − c_{1})(c_{2} − b_{2})d_{ev}(c_{2} − b_{2})\sqrt[∗]{c_{2} − b_{2}}
= (a_{2} + b_{2} − c_{2})(c_{1} − b_{1})d_{ev}(c_{1} − b_{1})\sqrt[∗]{c_{1} − b_{1}}.

(The form of R(α) obtained from R, where k_{j} = k(a_{j}, b_{j}, c_{j}) and n_{j} = n(a_{j}, b_{j}, c_{j}), for j = 1, 2, are defined in Theorem 6.) The equivalence classes determined by R and R(α), respectively, we denote by [ ]_{R} and [ ]_{R(α)}, respectively. From Theorem 6 it follows that the function α̂: ℤ × (ℤ\{0}) → P^{∗} of the form α̂([(k, n)]_{R}) = [α(k, n)]_{R(n)} is bijective. Since (ℤ × (ℤ\{0}))/R possesses the field structure (of fractions), α̂ transfers this structure onto P^{∗} automatically. We also have the (additive zero) element [0, 1]_{R} going to [(0, 0, 0)]_{R(α)} and the (multiplicative unit) element [(1, 1)]_{R} going to [1_{P}]_{R(α)}. It is easy to check that (a, b, c)R(α)(0, 0, 0) provided that a+b = c, and that (a, b, c)R(α)1_{P} whenever a + b − c = (c − b)d_{ev}(c − b)\sqrt[∗]{c − b} (for example, [1_{P}]_{R(α)} = [(6, 8, 10)]_{R(α)}(= [α(2, 2)]_{R(α)} = [(21, 72, 75)]_{R(α)}(= [α(3, 3)]_{R(α)}))). This proves the following theorem.

\underline{Theorem 9}. The set P^{∗}/R(α) is a commutative field under the following operations of addition ⊕ and multiplication ⊙

\begin{matrix}[(a_{1}, b_{1}, c_{1})]_{R(α)} ⊕ [(a_{2}, b_{2}, c_{2})]_{R(α)} := [α(k_{1}n_{2} + k_{2}n_{1}, n_{1}n_{2})]_{R(α)}, \\ [(a_{1}, b_{1}, c_{1})]_{R(α)} ⊙ [(a_{2}, b_{2}, c_{2})]_{R(α)} := [α(k_{1}k_{2}, n_{1}n_{2})]_{R(α)}, \end{matrix}

where for the given representatives (a′_{j}, b′_{j}, c′_{j}) of [(a_{j}, b_{j}, c_{j})]_{R(α)}, the integers k_{j} = k(a′_{j}, b′_{j}, c′_{j}) and n_{j} = n(a′_{j}, b′_{j}, c′_{j}), for j = 1, 2, are defined as in Theorem 6. The additive zero is [(0, 0, 0)]_{R(α)}, and the multiplicative unit is [(3, 4, 5)]_{R(α)}. The additive inverse of the element [(a, b, c)]_{R(α)} is

\begin{matrix}⊖ [(a, b, c)]_{R(α)} = [(A, B, C)]_{R(α)}, where \\ A = −\frac{a + b − c}{d_{ev}(c − b)} + c − b, B = (A^{2} − (c − b)^{2})/2(c − b), C = B + c − b \end{matrix}

(equivalently, ⊖[α(k, n)]_{R(α)} = [α(−k, n)]_{R(α)} ). The multiplicative inverse of the element [(a, b, c)]_{R(α)} for a + b ≠ c is

[(a, b, c)]_{R(α)}^{−1} = [(E, F, G)]_{R(α)}, where

E = (c − b) · d_{ev}(\frac{a + b − c}{d_{ev}(c − b) · \sqrt[∗]{c − b}}) · \sqrt[∗]{\frac{a + b − c}{d_{ev}(c − b) · \sqrt[∗]{c − b}}},

F = (E^{2} − (c − b)^{2})/2(c − b), G = F + c − b (equivalently, for k ≠ 0, we have [α(k, n)]_{R(α)}^{−1} = [α(n, k)]_{R(α)}].

By way of example, we have ⊖[(4, 3, 5)]_{R(α)} = ⊖[α(1, 2)]_{R(α)} = [α(−1, 2)]_{R(α)} = [(0, −1, 1)]_{R(α)}, and [(4, 3, 5)]_{R(α)}^{−1} = [α(1, 2)]_{R(α)}^{−1} = [α(2, 1)]_{R(α)} = [(5, 12, 13)]_{R(α)}.

\underline{References}

1. B. Dawson, “A Ring of Pythagorean Triples,” Missouri Journal of Mathematical Sciences, 6 (1994), 72–77.

2. A. Grytczuk, “Note on a Pythagorean Ring,” Missouri Journal of Mathematical Sciences, 9 (1997), 83–89.

3. M. Wójtowicz, “Algebraic Structures of Some Sets of Pythagorean Triples I,” Missouri Journal of Mathematical Sciences, 12 (2000), 31–35.

Marek Wójtowicz T. Kotarbinski Pedagogical University Institute of Mathematics 65-069 Zielona Góra, Poland email: mwojt@lord.wsp.zgora.pl