A NEW METHOD TO OBTAIN PYTHAGOREAN TRIPLE PRESERVING MATRICES

Mircea Crâşmăreanu

Abstract. Another method to obtain Pythagorean Triple Preserving Matrices is proposed and a singular case is put in evidence. Also, a possible connection with physics is sketched by proving that the set of these matrices is a group. In the last section, we generalize our method to Weighted Pythagorean Triple Preserving Matrices. An interesting open problem is generated by the fact that this type of matrix appears as a product of two matrices of order 4 with a form suggesting quaternions.

1. Pythagorean Triple Preserving Matrices. In [2] Palmer, Ahuja, and Tikoo obtained all matrices which convert a Pythagorean Triple into another Pythagorean Triple. In this paper we give a second method which uses the matrix equation of a quadric in real 3-dimensional space.

Recall that a Pythagorean Triple (PT) is a triple (a, b, c) of natural numbers such that a^{2} + b^{2} = c^{2} and recall that the general expression of a PT is

(a, b, c) = (m^{2} − n^{2}, 2mn, m^{2} + n^{2})

where m and n are two integers. So, a PT represents the coordinates of a point X ∈ ℝ^{3} which belongs to the quadric Γ : x^{2} + y^{2} − z^{2} = 0. The matrix equation of this quadric is Γ : X^{t} · S · X = 0 where

X = ( \begin{matrix}x \\ y \\ z \end{matrix} ) and S = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & −1 \end{matrix} ) .

Using [2], define a Pythagorean Triple Preserving Matrix (PTPM)

A = ( \begin{matrix}α_{1} & α_{2} & α_{3} \\ β_{1} & β_{2} & β_{3} \\ γ_{1} & γ_{2} & γ_{3} \end{matrix} ) .

That is, if X ∈ Γ then A · X ∈ Γ. Therefore, (AX)^{t} · S · (AX) = 0 which means X^{t} · (A^{t}SA) · X = 0. In conclusion, A is a PTPM if and only if there exists a real number ρ such that

A^{t}SA = ρS.  (1.1)

A straightforward computation leads to the following form of (1.1).

\{ \begin{matrix}α_{1}^{2} + β_{1}^{2} − γ_{1}^{2} = ρ \\ α_{2}^{2} + β_{2}^{2} − γ_{2}^{2} = ρ \\ α_{3}^{2} + β_{3}^{2} − γ_{3}^{2} = −ρ \\ α_{1}α_{2} + β_{1}β_{2} − γ_{1}γ_{2} = 0 \\ α_{2}α_{3} + β_{2}β_{3} − γ_{2}γ_{3} = 0 \\ α_{3}α_{1} + β_{3}β_{1} − γ_{3}γ_{1} = 0. \end{matrix}  (1.2)

If we make exactly the choice of [2], namely

\{ \begin{matrix}r^{2} = \frac{α_{1}+α_{3}+γ_{1}+γ_{3}}{2} ,  s^{2} = \frac{α_{3}−α_{1}+γ_{3}−γ_{1}}{2} \\ t^{2} = \frac{γ_{1}+γ_{3}−(α_{1}+α_{3})}{2},  u^{2} = \frac{γ_{3}−γ_{1}−(α_{3}−α_{1})}{2}. \end{matrix}  (1.3)

then, from (1.2_{1}), (1.2_{3}) and (1.2_{6}) it follows that

\{ \begin{matrix}β_{1} + β_{3} = 2rt \\ −β_{1} + β_{3} = 2su \end{matrix}

which gives

\{ \begin{matrix}β_{1} = rt − su \\ β_{3} = rt + su. \end{matrix}  (1.4)

From (1.3) we have, exactly as in [2], that

\{ \begin{matrix}α_{1} = \frac{(r^{2}−t^{2})−(s^{2}−u^{2})}{2},  α_{3} = \frac{(r^{2}−t^{2})+(s^{2}−u^{2})}{2} \\ γ_{1} = \frac{(r^{2}+t^{2})−(s^{2}+u^{2})}{2},  γ_{3} = \frac{(r^{2}+t^{2})+(s^{2}+u^{2})}{2} \end{matrix}  (1.5)

and then, from (1.2_{1}) it follows that

ρ = (ru − st)^{2} .  (1.6)
Equations (1.2_{2}), (1.2_{4}) and (1.2_{5}) yield

\{ \begin{matrix}α_{2} = rs − tu \\ β_{2} = ru + st \\ γ_{2} = rs + tu. \end{matrix}  (1.7)

In conclusion, from (1.4), (1.5) and (1.7), it follows that the general form of a PTPM is

A (r, s, t, u) = ( \begin{matrix}\frac{1}{2} (r^{2} − t^{2} − s^{2} + u^{2}) & rs − tu & \frac{1}{2} (r^{2} − t^{2} + s^{2} − u^{2}) \\ rt − su & ru + st & rt + su \\ \frac{1}{2} (r^{2} + t^{2} − s^{2} − u^{2}) & rs + tu & \frac{1}{2} (r^{2} + t^{2} + s^{2} + u^{2}) \end{matrix} ) (1.8)

which is exactly the expression given in [2].

A first advantage of the present method (which is of geometrical nature, like PT) is that it uses only 10 variables, namely (α_{i}) , (β_{i}) , (γ_{i}) and ρ, instead of 11 variables (α_{i}) , (β_{i}) , (γ_{i}) , M, N as in [2]. A second advantage is that given in the singular case ρ = 0 for relation (1.1) which we will discuss below. A third advantage is that it offers a very quick proof that the set of PTPM, considered with rational entries, is a group with respect to multiplication (see section 3).

We can obtain the pair (A (r, s, t, u) , ρ) from the product of two matrices of order 4. Considering

Φ_{1} = ( \begin{matrix}r & −s & −t & u \\ t & −u & r & −s \\ r & −s & t & −u \\ t & −u & −r & s \end{matrix} ) and Φ_{2} = ( \begin{matrix}r & s & r & s \\ s & −r & −s & r \\ t & u & t & u \\ u & −t & −u & t \end{matrix} )  (1.9)

we obtain

\frac{1}{2}Φ_{1} · Φ_{2} = \frac{1}{2} ( \begin{matrix}r^{2} − s^{2} − t^{2} + u^{2} & 2 (rs − tu) & r^{2} + s^{2} + t^{2} + u^{2} & 0 \\ 2 (rt − su) & 2 (ru + ts) & 2 (rt + su) & 0 \\ r^{2} − s^{2} + t^{2} − u^{2} & 2 (rs + tu) & r^{2} + s^{2} + t^{2} + u^{2} & 0 \\ 0 & 0 & 0 & −2 (ru − st) \end{matrix} )

= ( \begin{matrix}A (r, s, t, u) & 0 \\ 0 & −\sqrt{ρ} \end{matrix} )  (1.10)

and this fact, using the expression of Φ_{1} and Φ_{2} yields the following problem.

\underline{Open problem}. Does there exist a connection between PTPM and the algebra of quaternions?

As a possible answer, let us note that the matrix (1.8) is close to the matrix from [4] representing the rotations in ℝ^{3}.

2. The Singular Case. For relation (1.1) the case ρ = 0 appears as a singular case. From relation (1.6) we have ru = st.

Case I. Suppose that one of r or u is zero. Then one of t and s is zero. We make the choice r = s = 0 and then it follows that

A (0, 0, t, u) = ( \begin{matrix}\frac{u^{2}−t^{2}}{2} & −tu & \frac{−u^{2}−t^{2}}{2} \\ 0 & 0 & 0 \\ \frac{t^{2}−u^{2}}{2} & tu & \frac{t^{2}+u^{2}}{2} \end{matrix} ) = \frac{t^{2}}{2} ( \begin{matrix}−1 & 0 & −1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{matrix} )

+ \frac{u^{2}}{2} ( \begin{matrix}1 & 0 & −1 \\ 0 & 0 & 0 \\ −1 & 0 & 1 \end{matrix} ) + tu ( \begin{matrix}0 & −1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} )

= t^{2}A (0, 0, 1, 0) + u^{2}A (0, 0, 0, 1) + tu ( \begin{matrix}0 & −1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) . (2.1)

We have

A (0, 0, 1, 0) ( \begin{matrix}m^{2} − n^{2} \\ 2mn \\ m^{2} + n^{2} \end{matrix} ) = ( \begin{matrix}−m^{2} \\ 0 \\ m^{2} \end{matrix} ) and A (0, 0, 0, 1) ( \begin{matrix}m^{2} − n^{2} \\ 2mn \\ m^{2} + n^{2} \end{matrix} ) = ( \begin{matrix}−n^{2} \\ 0 \\ n^{2} \end{matrix} ) .

That is, we obtain the “singular” PT (−1, 0, 1) .

Case II. Suppose that r, u ≠ 0. Then s, t ≠ 0. From the relation s = \frac{ru}{t} it follows that

A (r, \frac{ru}{t}, t, u) = ( \begin{matrix}\frac{1}{2} (r^{2} − t^{2} − \frac{r^{2}u^{2}}{t^{2}} + u^{2}) & \frac{r^{2}u}{t} − tu & \frac{1}{2} (r^{2} − t^{2} + \frac{r^{2}u^{2}}{t^{2}} − u^{2}) \\ rt − \frac{ru^{2}}{t} & 2ru & rt + \frac{ru^{2}}{t} \\ \frac{1}{2} (r^{2} + t^{2} − \frac{r^{2}u^{2}}{t^{2}} − u^{2}) & \frac{r^{2}u}{t} + tu & \frac{1}{2} (r^{2} + t^{2} + \frac{r^{2}u^{2}}{t^{2}} + u^{2}) \end{matrix} ) .

(2.2)

For example,

A (r, ru, 1, u) = ( \begin{matrix}\frac{1}{2} (r^{2} − 1 − r^{2}u^{2} + u^{2}) & u (r^{2} − 1) & \frac{1}{2} (r^{2} − 1 + r^{2}u^{2} − u^{2}) \\ r (1 − u^{2}) & 2ru & r (1 + u^{2}) \\ \frac{1}{2} (r^{2} + 1 − r^{2}u^{2} − u^{2}) & u (r^{2} + 1) & \frac{1}{2} (r^{2} + 1 + r^{2}u^{2} + u^{2}) \end{matrix} )

(2.3)

and then

A (r, r, 1, 1) = ( \begin{matrix}0 & r^{2} − 1 & r^{2} − 1 \\ 0 & 2r & 2r \\ 0 & r^{2} + 1 & r^{2} + 1 \end{matrix} )  (2.4)

which gives

A (r, r, 1, 1) ( \begin{matrix}m^{2} − n^{2} \\ 2mn \\ m^{2} + n^{2} \end{matrix} ) = ( \begin{matrix}(r^{2} − 1) & (m + n) & 2 \\ 2r (m + n)^{2} \\ (r^{2} + 1) & (m + n) & 2 \end{matrix} ) .

So,

A (1, 1, 1, 1) ( \begin{matrix}m^{2} − n^{2} \\ 2mn \\ m^{2} + n^{2} \end{matrix} ) = ( \begin{matrix}0 \\ 2 (m + n)^{2} \\ 2 (m + n)^{2} \end{matrix} )

i.e. the “singular” PT (0, 1, 1). Comparing the results of this section with a proposition from [2]: “no specific conditions on the nature of r, s, t and u are imposed”. In conclusion, relation (1.1) characterizes PTPM yielding “non-singular” PT only for the case ρ ≠ 0.

3. Connections With Physics. A field of possible applications for the previous results is the 2 + 1 Theory of Relativity. Consider ℝ^{3} with the Lorentzian metric ([5])

< A⃗, B⃗ >= a_{1}b_{1} + a_{2}b_{2} − a_{3}b_{3}  (3.1)

for A⃗ = (a_{1}, a_{2}, a_{3}), B⃗ = (b_{1}, b_{2}, b_{3}) ∈ ℝ^{3}. In [5], the pair E^{3,1} = (ℝ^{3}, <, >) is called the Minkowski 3-space. In this space-time the quadric Γ : x^{2} + y^{2} − z^{2} = 0 is exactly the set of null vectors ([5]). More precisely, Γ is the null cone of E^{3,1} because if A⃗ ∈ Γ then λA⃗ ∈ Γ for all real λ.

Therefore, a PT represents a point in the null cone, with natural coordinates and then a PTPM is a linear transformation of E^{3,1} which preserves the points of natural coordinates from the null cone of E^{3,1}.

Using (1.1) results in the fact that the set of PTPM with rational entries is a group with respect to multiplication. Indeed, A is the unit matrix for r = u = 1, s = t = 0 and if A_{1} and A_{2} are PTPM with corresponding ρ_{1} and ρ_{2}, then (1.1) yields

(A_{1}A_{2})^{t} S (A_{1}A_{2}) = A_{2}^{t} (A_{1}^{t}SA_{1}) A_{2} = ρ_{1}A_{2}^{t}SA_{2} = ρ_{1}ρ_{2}S

which means that A_{1}A_{2} is a PTPM with corresponding ρ_{1}ρ_{2}. With MAPLE it is easy to obtain the relation

A (r_{1}, s_{1}, t_{1}, u_{1}) · A (r_{2}, s_{2}, t_{2}, u_{2})
= A (r_{1}r_{2} + t_{2}s_{1}, r_{1}s_{2} + u_{2}s_{1}, r_{2}t_{1} + t_{2}u_{1}, t_{1}s_{2} + u_{1}u_{2})  (3.2)

which implies

A^{2} (r, s, t, u) = A (r^{2} + ts, (r + u) s, (r + u) t, ts + u^{2})  (3.3)

A^{−1} (r, s, t, u) = A (\frac{u}{ru − st}, \frac{−s}{ru − st}, \frac{−t}{ru − st}, \frac{r}{ru − st})  (3.4)

for ru ≠ st (see the previous section). Other properties of A (r, s, t, u) which are obtained with MAPLE are

(i) The trace is

T rA = r^{2} + u^{2} + ru + st.  (3.5)

(ii) The eigenvalues are

λ_{1} = ru − st  (3.6a)
λ_{2} = \frac{1}{2} (r^{2} + u^{2}) + ts + \frac{1}{2}\sqrt{r^{4} − 2r^{2}u^{2} + 4r^{2}st + u^{4} + 4u^{2}st + 8rstu} (3.6b)

λ_{3} = \frac{1}{2} (r^{2} + u^{2}) + ts − \frac{1}{2}\sqrt{r^{4} − 2r^{2}u^{2} + 4r^{2}st + u^{4} + 4u^{2}st + 8rstu}. (3.6c)

4. Weighted Pythagorean Triple Preserving Matrices. A Weighted Pythagorean Triple (WPT) is a triple (x, y, z) of natural numbers such that

p^{2}x^{2} + q^{2}y^{2} = p^{2}q^{2}z^{2}  (4.1)

where p and q are two natural numbers. So, a WPT represents the coordinates of a point X ∈ ℝ^{3} which belongs to the quadric Γ : p^{2}x^{2} + q^{2}y^{2} − p^{2}q^{2}z^{2} = 0. The matrix equation of this quadric is Γ : X^{t} · S(p, q) · X = 0 where

X = ( \begin{matrix}x \\ y \\ z \end{matrix} ) and S(p, q) = ( \begin{matrix}p^{2} & 0 & 0 \\ 0 & q^{2} & 0 \\ 0 & 0 & −p^{2}q^{2} \end{matrix} ) .

In this section we find the general form of a Weighted Pythagorean Triple Preserving Matrix (WPTPM) A, i.e., if X ∈ Γ then A · X ∈ Γ. Using the same argument as in the first section results in the fact that A is a WPTPM if and only if there exists a real number ρ such that

A^{t} · S(p, q) · A = ρS (p, q) .  (4.2)

A straightforward computation leads to the following form of (4.2).

\{ \begin{matrix}p^{2}α_{1}^{2} + q^{2}β_{1}^{2} − p^{2}q^{2}γ_{1}^{2} = ρp^{2} \\ p^{2}α_{2}^{2} + q^{2}β_{2}^{2} − p^{2}q^{2}γ_{2}^{2} = ρq^{2} \\ p^{2}α_{3}^{2} + q^{2}β_{3}^{2} − p^{2}q^{2}γ_{3}^{2} = −ρp^{2}q^{2} \\ p^{2}α_{1}α_{2} + q^{2}β_{1}β_{2} − p^{2}q^{2}γ_{1}γ_{2} = 0 \\ p^{2}α_{2}α_{3} + q^{2}β_{2}β_{3} − p^{2}q^{2}γ_{2}γ_{3} = 0 \\ p^{2}α_{3}α_{1} + q^{2}β_{3}β_{1} − p^{2}q^{2}γ_{3}γ_{1} = 0. \end{matrix}  (4.3)

With the choice

\{ \begin{matrix}r^{2} = \frac{1}{2q^{2}} [pq (γ_{3} + qγ_{1}) + p (α_{3} + qα_{1})] , s^{2} = \frac{1}{2q^{2}} [pq (γ_{3} − qγ_{1}) + p (α_{3} − qα_{1})] \\ t^{2} = \frac{1}{2q^{2}} [pq (γ_{3} + qγ_{1}) − p (α_{3} + qα_{1})] , u^{2} = \frac{1}{2q^{2}} [pq (γ_{3} − qγ_{1}) − p (α_{3} − qα_{1})] \end{matrix}
(4.4)

it follows that the solution

A (r, s, t, u) = ( \begin{matrix}\frac{q}{2p} (r^{2} − t^{2} − s^{2} + u^{2}) & \frac{q^{2}}{p^{2}} (rs − tu) & \frac{q^{2}}{2p} (r^{2} − t^{2} + s^{2} − u^{2}) \\ rt − su & \frac{q}{p} (ru + st) & q (rt + su) \\ \frac{1}{2p} (r^{2} + t^{2} − s^{2} − u^{2}) & \frac{q}{p^{2}} (rs + tu) & \frac{q}{2p} (r^{2} + t^{2} + s^{2} + u^{2}) \end{matrix} ) .

(4.5)

Also,

ρ = \frac{q^{2}}{p^{2}} (ru − st)^{2} .  (4.6)

Returning to (4.1) with x = qa and y = pb results in a^{2} + b^{2} = z^{2}, i.e. (a, b, z) is a PT and therefore, we have the general form of a WPT.

(x, y, z) = (q (m^{2} − n^{2}) , 2pmn, m^{2} + n^{2}) .  (4.7)

Finally, consider the system

A (r, s, t, u) · ( \begin{matrix}q (m^{2} − n^{2}) \\ 2pmn \\ m^{2} + n^{2} \end{matrix} ) = ( \begin{matrix}q (M^{2} − N^{2}) \\ 2pMN \\ M^{2} + N^{2} \end{matrix} )  (4.8)

with solution

M^{2} = \frac{q}{p} (mr + ns)^{2} , N^{2} = \frac{q}{p} (mt + ns)^{2} .  (4.9)

This yields the following proposition.

\underline{Proposition}. The results of this section are true only for the case

q = p · α^{2}  (4.10)

with α a natural number. Then (4.1) becomes

x^{2} + α^{4}y^{2} = p^{2}α^{4}z^{2}.  (4.11)

Obviously, for p = q = 1 we reobtain the results of the first section.

\underline{References}

1. R. Miron and D. Brânzei, Backgrounds of Arithmetic and Geometry. An Introduction, Series in Pure Mathematics, no. 23, World Scientific, 1995.

2. L. Palmer, M. Ahuja, and M. Tikoo, “Finding Pythagorean Triple Preserving Matrices,” Missouri Journal of Mathematical Sciences, 10 (1998), 99–105.

3. L. Palmer, M. Ahuja, and M. Tikoo, “Constructing Pythagorean Triple Preserving Matrices,” Missouri Journal of Mathematical Sciences, 10 (1998), 159– 168.

4. J. P. Ward, Quaternions and Cayley Numbers. Algebra and Applications, Mathematics and Its Applications, no. 403, Kluwer Academic Publishers, 1997.

5. T. Weinstein, An Introduction to Lorentz Surfaces, W. de Gruyter Expositions in Mathematics, no. 22, 1996.

Mircea Crâşmăreanu Faculty of Mathematics University “Al. I. Cuza” Iaşi, 6600, Romania email: mcrasm@uaic.ro

Institute of Mathematics “Octav Mayer” Iasi Branch of Romanian Academy Iaşi, 6600, Romania