MONOIDS CONNECTED WITH EULER’S DIOPHANTINE EQUATION

Aleksander Grytczuk

Abstract. In this paper we give a construction of infinite monoids generated by the integer solutions of Euler’s Diophantine equation x^{2} + y^{2} = z^{n}, n ≥ 2.

1. Introduction. It is well–known, by the classical result of Euler, that the Diophantine equation

x^{2} + y^{2} = z^{n}, n ≥ 2  (1)

has infinitely many solutions in integers x, y, and z for any fixed positive integer n ≥ 2. Moreover, all integer solutions of (1) in integers x, y, and z such that (x, y) = 1 are given by the following formula:

x = \frac{(r + is)^{n} + (r − is)^{n}}{2} , y = \frac{(r + is)^{n} − (r − is)^{n}}{2i} , z = r^{2} + s^{2},  (2)

where r and s are integers such that (r, s) = 1. Let

S_{n} = {⟨x, y, z⟩ ∈ ℤ^{3}; x^{2} + y^{2} = z^{n}; n ≥ 2}.  (3)

Define the operation “∘” on S_{n} as follows. If α = ⟨a, b, c⟩ ∈ S_{n} and β = ⟨u, v, w⟩ ∈ S_{n}, then

α ∘ β = ⟨a, b, c⟩ ∘ ⟨u, v, w⟩ = ⟨au − bv, av + bu, cw⟩ = γ.  (4)

The following identity

(au − bv)^{2} + (av + bu)^{2} = (a^{2} + b^{2})(u^{2} + v^{2})  (5)

is well–known. Since α, β ∈ S_{n}, then by (3) it follows that

a^{2} + b^{2} = c^{n}, u^{2} + v^{2} = w^{n}.  (6)

From (5) and (6), we obtain

(au − bv)^{2} + (av + bu)^{2} = (cw)^{n}.  (7)

Now, by (7), it follows that the element γ = ⟨au − bv, av + bu, cw⟩ belongs to S_{n}. We prove that the set ⟨S_{n}; ∘⟩, where the operation “∘” is define by (4) is a commutative monoid for any fixed positive integer n ≥ 2. We note that in the case n = 2, the equation (1) reduces to the Pythagorean equation. In this case, B. Dawson [1] gave a construction of a Pythagorean ring. He defined two operations and an isomorphism Φ: P → ℤ × ℤ, where P = {⟨x, y, z⟩ ∈ ℤ^{3}; x^{2} + y^{2} = z^{2}} and utilizing the elements of the set P_{n} = {⟨x, y, z⟩ ∈ P; z − y = n}. Moreover, in [2], it was proven that the set P_{n}, with respect to the particular operations “⊕” and “∘” is a commutative ring for any fixed integer n.

2. Results. We begin by proving the following theorem.

\underline{Theorem 1}. The set ⟨S_{n}; ∘⟩, where the operation “∘” is defined by (4) and S_{n} by (3), is a commutative monoid, for any positive integer n ≥ 2.

\underline{Proof}. Let α = ⟨a, b, c⟩ ∈ S_{n}, β = ⟨u, v, w⟩ ∈ S_{n}, and γ = ⟨d, e, f⟩ ∈ S_{n}. Then by (4), it follows that

\begin{aligned}L & = (α ∘ β) ∘ γ = (⟨a, b, c⟩ ∘ ⟨u, v, w⟩) ∘ ⟨d, e, f⟩ \\ & = ⟨au − bv, av + bu, cw⟩ ∘ ⟨d, e, f⟩. \end{aligned}

Putting a_{1} = au − bv, b_{1} = av + bu, and c_{1} = cw in the last equality and using (4), we obtain

L = ⟨a_{1}, b_{1}, c_{1}⟩ ∘ ⟨d, e, f⟩ = ⟨a_{1}d − b_{1}e, a_{1}e + b_{1}d, c_{1}f⟩.

In a similar way, we obtain

\begin{aligned}P & = α ∘ (β ∘ γ) = ⟨a, b, c⟩ ∘ (⟨u, v, w⟩ ∘ ⟨d, e, f⟩) \\ & = ⟨a, b, c⟩ ∘ ⟨ud − ve, ue + vd, wf⟩. \end{aligned}

Let u_{1} = ud − ve, v_{1} = ue + vd, and w_{1} = wf. Then by (4) and the last equality, it follows that

P = ⟨a, b, c⟩ ∘ ⟨u_{1}, v_{1}, w_{1}⟩ = ⟨au_{1} − bv_{1}, av_{1} + bu_{1}, cw_{1}⟩.

Moreover, it is easy to see that

cw_{1} = cwf = c_{1}f  (8)
a_{1}d − b_{1}e = (au − bv)d − (av + bu)e = a(ud − ve) − b(ue + vd) = av_{1} + bu_{1} (9)
a_{1}e + b_{1}d = (au − bv)e + (av + bu)d = a(ue + vd) + b(ud − ve) = av_{1} + bu_{1}. (10)

From (8)–(10), it follows that L = P and the associative law is satisfied. On the other hand, we have

α ∘ β = ⟨a, b, c⟩ ∘ ⟨u, v, w⟩ = ⟨au − bv, av + bu, cw⟩ = β ∘ α

and so the commutative law is satisfied. Finally, we remark that the element e = ⟨1, 0, 1⟩ ∈ S_{n} and for every α ∈ S_{n} we have, by (4), that α ∘ e = e ∘ α = α; therefore, the element e = ⟨1, 0, 1⟩ is the identity element in the set S_{n}, and the proof of Theorem 1 is complete.

\underline{Remark}. The set ⟨S_{n}; ∘⟩ is not a group, because α has an inverse in S_{n} if and only if α = ⟨±1, 0, ±1⟩ or ⟨0, ±1, ±1⟩.

Now, we introduce a special set of matrices:

M_{2}^{(n)}(ℤ) = \{A = ( \begin{matrix}a & b \\ −b & a \end{matrix} ); det A = c^{n}; n ≥ 2; a, b, c ∈ ℤ\}.  (∗)

We prove the following.

\underline{Theorem 2}. Let M_{2}^{(n)}(ℤ) be the set of all integral matrices defined by (∗) with the operation of matrix multiplication, denoted by “·”. Then the set ⟨S_{n}; ∘⟩ is isomorphic to the set ⟨M_{2}^{(n)}(ℤ), ·⟩.

\underline{Proof}. Let Φ: S_{n} → M_{2}^{(n)}(ℤ) be the mapping defined as follows. If α = ⟨a, b, c⟩ ∈ S_{n}, then

Φ(α) = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) = A; det A = c^{n}; n ≥ 2.  (11)

First, we remark that the mapping Φ is bijective. Further, for α = ⟨a, b, c⟩ ∈ S_{n} and β = ⟨u, v, w⟩ ∈ S_{n}, we have

Φ(α ∘ β) = Φ(⟨a, b, c⟩ ∘ ⟨u, v, w⟩) = Φ(⟨au − bv, av + bu, cw⟩).

From the last equality and (11), we obtain

Φ(α ∘β) = ( \begin{matrix}au − bv & av + bu \\ −(av + bu) & au − bv \end{matrix} ) = C, det C = (cw)^{n} = (au−bv)^{2} +(av +bu)^{2}.

On the other hand, by (11), it follows that

Φ(α) = Φ(⟨a, b, c⟩) = A = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) , det A = a^{2} + b^{2} = c^{n}  (12)

Φ(β) = Φ(⟨u, v, w⟩) = B = ( \begin{matrix}u & v \\ −v & u \end{matrix} ) , det B = u^{2} + v^{2} = w^{n}.  (13)

By (12) and (13), it follows that

Φ(α) · Φ(β) = A · B = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) · ( \begin{matrix}u & v \\ −v & u \end{matrix} ) = ( \begin{matrix}au − bv & av + bu \\ −(av + bu) & au − bv \end{matrix} ) . (14)

For (14) and Cauchy’s theorem on the product of determinants, we obtain

det(A · B) = det A · det B = (a^{2} + b^{2})(u^{2} + v^{2}) = (au − bv)^{2} + (av + bu)^{2}. (15)

By (12), (13), and (15), it follows that (cw)^{n} = (au − bv)^{2} + (av + bu)^{2} = det C and consequently, we obtain Φ(α ∘ β) = Φ(α) · Φ(β), hence, S_{n} ≈ M_{2}^{(n)}(ℤ). The proof of Theorem 2 is complete.

\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.

Aleksander Grytczuk Institute of Mathematics Department of Algebra and Number Theory T. Kotarbiński Pedagogical University 65-069 Zielona Góra, Poland email: grytal@omega.im.wsp.zgora.pl