SYMMETRIC PYTHAGOREAN TRIPLE PRESERVING MATRICES

James T. Bruening, Tracy R. Lohmeier, and Cheryl L. Sebaugh

Abstract. A Pythagorean Triple Preserving Matrix (PTPM) is a 3×3 matrix such that if it is multiplied by a Pythagorean Triple, the result is also a Pythagorean Triple. Necessary and sufficient conditions for a Pythagorean Triple Preserving Matrix to be symmetric are given. Monoids of Symmetric Pythagorean Triple Preserving Matrices (SPTPM) with positive integer entries will be developed, and to ensure that the set is closed under matrix multiplication, the focus will be on finding commutative SPTPM’s.

1. Introduction. A triple of positive integers (a, b, c) is defined to be a Pythagorean Triple if it satisfies the equation a^{2} + b^{2} = c^{2}. Moreover, (a, b, c) is said to be a Primitive Pythagorean Triple (PPT) if gcd(a, b, c) = 1 [3].

It is known [1] that all Primitive Pythagorean Triples (a, b, c) are of the form (a, b, c) = (m^{2} − n^{2}, 2mn, m^{2} + n^{2}) where m, n satisfy the conditions listed below:

I.1 m, n are positive integers.

I.2 m > n.

I.3 gcd(m, n) = 1. I.4 m ≢ n (mod 2).

It should be noted that a triple (m^{2} − n^{2}, 2mn, m^{2} + n^{2}) satisfies the Pythagorean Theorem for every value of m and n. However, only those that satisfy conditions I.1 to I.4 above are called PPT’s. Also, this form implies that if (a, b, c) is a Primitive Pythagorean Triple, then b must be even.

A 3×3 matrix A with integer entries is called a Pythagorean Triple Preserving Matrix if whenever (a, b, c) is a PT, then (d, e, f) = (a, b, c)A is also a PT [1]. For example, let

A = [ \begin{matrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{matrix} ] .

One can easily see that the Pythagorean Triple (3, 4, 5) multiplied by A gives (3, 4, 5)A = (21, 20, 29), which is yet another Pythagorean Triple [4]. In two recent papers by Palmer, Ahuja, and Tikoo [1,2], Pythagorean Triple Preserving Matrices have been identified and constructed.

The following lemma and theorem are proven by Palmer, Ahuja, and Tikoo [1].

\underline{Lemma 1}. Let A denote the matrix

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

Then (m^{2} −n^{2}, 2mn, m^{2} +n^{2})A = (M^{2} −N^{2}, 2MN, M^{2} +N^{2}), where M = mr+nt, N = ms + nu. The relation between (m, n) and (M, N) can be expressed by the matrix equation

(m, n) [ \begin{matrix}r & s \\ t & u \end{matrix} ] = (M, N).  (2)

\underline{Theorem 1}. A 3 × 3 matrix is a PTPM if and only if it is of form A.

The authors in [1] further show that the values of r, s, t, u should be restricted in the following ways:

R-1. r, s, t, u are integers with r, s > 0. R-2. r + t ≥ s + u ≥ 0. (Later in this paper, it will be further assumed that r > u.) R-3. ru − st = ±1. R-4. r + s ≡ t + u ≡ 1 (mod 2).

These restrictions are sufficient to assure that the second triple (M^{2} − N^{2}, 2MN, M^{2} + N^{2}) is a PPT whenever (m^{2} − n^{2}, 2mn, m^{2} + n^{2}) is a PPT. It is also shown in [1] that det A = (ru − st)^{3}, so since (±1)^{3} = ±1, only PTPM’s with determinant ±1 will be considered.

A semigroup is a nonempty set G together with an associative binary operation; a monoid is a semigroup G which contains a (two–sided) identity element e ∈ G such that ae = ea = a for all a ∈ G [6]. Therefore since (1) the identity is obviously a PTPM, (2) the set of PTPM’s is easily seen to be closed under the operation of matrix multiplication, and (3) matrix multiplication is associative, it is known that the set of PTPM’s forms a monoid.

A square matrix S is said to be symmetric if S = S^{T}, where S^{T} denotes the transpose of S. The set of all Symmetric Pythagorean Triple Preserving Matrices is not closed under the operation of matrix multiplication since the product of two symmetric matrices may or may not be symmetric. It can be easily shown, however, that the product of two symmetric matrices is symmetric if and only if the two matrices commute.

It is known for symmetric matrices [7,8] that any two eigenvectors from different eigenspaces are orthogonal. Furthermore, an (n × n) matrix A is orthogonally diagonalizable if and only if A is symmetric, and a matrix is known to be diagonalizable if and only if it possess a set of n linearly independent eigenvectors. An n×n matrix A is defined to be semisimple if it has a total of n linearly independent eigenvectors [5], so every symmetric matrix is semisimple. The following theorem [5] will then guarantee when the product of two SPTPM’s is symmetric.

\underline{Theorem 2}. If two matrices are semisimple, then they commute if and only if they have a complete set of eigenvectors in common.

2. Conditions That Guarantee Symmetry.

\underline{Theorem 3}. Let

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

Then A is symmetric if and only if s = t.

The proof follows in a straightforward manner.

\underline{Remark}. Note that if s = t, then the matrix

[ \begin{matrix}r & s \\ t & u \end{matrix} ]

in equation (2) is symmetric as well.

The symmetric PTPM, for the remainder of this paper, will be denoted J, where J has the form

J = [ \begin{matrix}\frac{r^{2}−2s^{2}+u^{2}}{2} & rs − su & \frac{r^{2}−u^{2}}{2} \\ rs − su & ru + s^{2} & rs + su \\ \frac{r^{2}−u^{2}}{2} & rs + su & \frac{r^{2}+2s^{2}+u^{2}}{2} \end{matrix} ] .

The symmetry condition s = t when applied to the determinant condition, R-3 (ru − st = ±1), produces the two equations, ru − s^{2} = 1 and ru − s^{2} = −1. The first of these equations can be rewritten ru = s^{2} + 1, but this rewritten form does not have any obvious solutions. The second equation, though, when rewritten ru = s^{2} − 1 = (s + 1)(s − 1) produces the natural solutions r = s + 1, u = s − 1. Substituting these into (1) generates symmetric PTPM’s of the form

C = [ \begin{matrix}1 & 2s & 2s \\ 2s & 2s^{2} − 1 & 2s^{2} \\ 2s & 2s^{2} & 2s^{2} + 1 \end{matrix} ] ,

where det C = (ru − s^{2}) = −1. The example

[ \begin{matrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{matrix} ]

illustrated earlier in this paper is generated by r = 2, s = t = 1, u = 0.

It is a routine matter to show that J is a symmetric PTPM of form C if and only if r = s + 1 and u = s − 1. The significance of matrices of form C will be illustrated in later sections when monoids of symmetric PTPM’s are studied.

3. Eigenvalues and Eigenvectors. All SPTPM’s are of the form

J = [ \begin{matrix}\frac{r^{2}−2s^{2}+u^{2}}{2} & rs − su & \frac{r^{2}−u^{2}}{2} \\ rs − su & ru + s^{2} & rs + su \\ \frac{r^{2}−u^{2}}{2} & rs + su & \frac{r^{2}+2s^{2}+u^{2}}{2} \end{matrix} ] ,

where r, s > 0, ru − s^{2} = ±1. The condition R-2 (r > u) guarantees the positive nature of the elements and the condition R-4 (r+s ≡ t+u ≡ 1 (mod 2)) guarantees that the entries are integers. Through the use of computer software, the eigenvalues λ_{i} and eigenvectors x_{i} of J were found:

\begin{matrix}λ_{1} = ru − s^{2} & x_{1} = [ \frac{2s}{u−r} ; 1 ; 0 ] \\ λ_{2} = \frac{r^{2}+2s^{2}+u^{2}}{2} + ^{\frac{|r+u|\sqrt{r^{2}−2ru+4s^{2}+u^{2}}}{2}} & x_{2} = [ 1 ; \frac{2s}{r−u} ; \frac{\sqrt{r^{2}−2ru+4s^{2}+u^{2}}}{r−u} ] \\ λ_{3} = \frac{r^{2}+2s^{2}+u^{2}}{2} + ^{\frac{|r+u|\sqrt{r^{2}−2ru+4s^{2}+u^{2}}}{2}} & x_{3} = [ 1 ; \frac{2s}{r−u} ; \frac{\sqrt{r^{2}−2ru+4s^{2}+u^{2}}}{u−r} ] \end{matrix} .

4. Conditions for SPTPM’s to Commute.

\underline{Theorem 4}. Let J_{1} and J_{2} be SPTPM’s generated by r_{1}, s_{1}, and u_{1}, and r_{2}, s_{2}, and u_{2}, respectively. Then J_{1} and J_{2} commute if and only if

\frac{s_{1}}{r_{1} − u_{1}} = \frac{s_{2}}{r_{2} − u_{2}} .

\underline{Proof}. Assume the semisimple matrices J_{1} and J_{2} commute. Let the eigenvectors be as follows:

J_{1} : x_{1} = [ \frac{2s_{1}}{u_{1}−r_{1}} ; 1 ; 0 ] , x_{2} = [ 1 ; \frac{2s_{1}}{r_{1}−u_{1}} ; \frac{\sqrt{r_{1}^{2}−2r_{1}u_{1}+4s_{1}^{2}+u_{1}^{2}}}{r_{1}−u_{1}} ] , x_{3} = [ 1 ; \frac{2s_{1}}{r_{1}−u_{1}} ; \frac{\sqrt{r_{1}^{2}−2r_{1}u_{1}+4s_{1}^{2}+u_{1}^{2}}}{u_{1}−r_{1}} ]

and

J_{2} : y_{1} = [ \frac{2s_{2}}{u_{2}−r_{2}} ; 1 ; 0 ] , y_{2} = [ 1 ; \frac{2s_{2}}{r_{2}−u_{2}} ; \frac{\sqrt{r_{2}^{2}−2r_{2}u_{2}+4s_{2}^{2}+u_{2}^{2}}}{r_{2}−u_{2}} ] , y_{3} = [ 1 ; \frac{2s_{2}}{r_{2}−u_{2}} ; \frac{\sqrt{r_{2}^{2}−2r_{2}u_{2}+4s_{2}^{2}+u_{2}^{2}}}{u_{2}−r_{2}} ] .

Since J_{1} and J_{2} must have a complete set of common eigenvectors by Theorem 2, the pairing x_{1} = y_{1}, x_{2} = y_{2}, x_{3} = y_{3}, is the only possibility. This results in the following conditions:

\frac{2s_{1}}{r_{1} − u_{1}} = \frac{2s_{2}}{r_{2} − u_{2}} ,  (3)

and

\frac{\sqrt{r_{1}^{2} − 2r_{1}u_{1} + 4s_{1}^{2} + u_{1}^{2}}}{r_{1} − u_{1}} = \frac{\sqrt{r_{2}^{2} − 2r_{2}u_{2} + 4s_{2}^{2} + u_{2}^{2}}}{r_{2} − u_{2}} .  (4)

It can be shown that (3) and (4) are equivalent. (Assume r > u.) This implies that when the eigenvectors are in common, then

\frac{s_{1}}{r_{1} − u_{1}} = \frac{s_{2}}{r_{2} − u_{2}} .

Conversely, assume

\frac{s_{1}}{r_{1} − u_{1}} = \frac{s_{2}}{r_{2} − u_{2}}

and let the eigenvectors of J_{1} and J_{2} be the same as above. Since equations (3) and (4) are equivalent, the eigenvectors of J_{1} and J_{2} are in common by the pairing x_{1} = y_{1}, x_{2} = y_{2}, x_{3} = y_{3}, and the matrices commute by Theorem 2.

\underline{Corollary 1}. Let C_{1} be a matrix of form C generated by s = k and let J_{1} be a matrix of form J generated by r_{1}, s_{1}, and u_{1}. C_{1} and J_{1} commute if and only if

k = \frac{2s_{1}}{r_{1} − u_{1}} .

\underline{Corollary 2}. If C_{1} and C_{2} are two SPTPM’s of form C, then C_{1} and C_{2} commute if and only if C_{1} = C_{2}.

\underline{Theorem 5}. If J_{1} and C_{1} are SPTPM’s of form J and C, respectively, then J_{1} and C_{1} commute if and only if J_{1} = C_{1}^{n}, n ≥ 0. (C_{1}^{0} = I_{3}, the 3 by 3 identity matrix.)

\underline{Proof}. Let J_{1} = C_{1}^{n}, n ≥ 0. Then J_{1} ·C_{1} = C_{1}^{n} ·C_{1} = C_{1}^{n+1} = C_{1} ·C_{1}^{n} = C_{1} ·J_{1}, and the matrices commute.

Now assume J_{1} and C_{1} commute, where

C_{1} = [ \begin{matrix}1 & 2k & 2k \\ 2k & 2k^{2} − 1 & 2k^{2} \\ 2k & 2k^{2} & 2k^{2} + 1 \end{matrix} ] ,

k a positive integer, and J_{1} is an SPTPM generated by r_{1}, s_{1}, u_{1}. From condition R-3,

r_{1}u_{1} − s_{1}^{2} = ±1,  (5)

and from Corollary 1,

k = \frac{2s_{1}}{r_{1} − u_{1}} .  (6)

Combining equations (5) and (6),

(\frac{2s_{1} + ku_{1}}{k} )u_{1} − s_{1}^{2} = ±1

ku_{1}^{2} + 2s_{1}u_{1} − ks_{2} = ±k ku_{1}^{2} + 2s_{1}u_{1} − ks_{2} ± k = 0.

The quadratic formula will then give

u_{1} = \frac{−s_{1} + \sqrt{s_{1}^{2}(1 + k^{2}) ± k^{2}}}{k} .  (7)

Similarly,

r_{1} = \frac{s_{1} + \sqrt{s_{1}^{2}(1 + k^{2}) ± k^{2}}}{k} .  (8)

The minus signs before the radicals in u_{1} and r_{1} from the quadratic formula are eliminated to guarantee r_{1} is positive, and so that equation (6) is satisfied.

From condition R-4, r_{1} − u_{1} must be even, and equation (6) implies

\frac{r_{1} − u_{1}}{2} = \frac{s_{1}}{k} ,

so that k divides s_{1}. Let s_{1} = q · k. Then \sqrt{s_{1}^{2}(k^{2} + 1) ± k^{2}} = k\sqrt{q^{2}(k^{2} + 1) ± 1}. If p^{2} = q^{2}(k^{2} + 1) ± 1, p an integer, then

p^{2} − q^{2}(k^{2} + 1) = ±1,  (9)

is a form of Pell’s equation, and solutions (p, q) can be found by methods outlined in number theory books [3,9]. It can be shown that all positive integer solutions of (9) are of the form (p_{n}, q_{n}), n ≥ 1, where p_{0} = 1, p_{1} = k, …, p_{n} = 2k·p_{n−1} +p_{n−2}, and q_{0} = 0, q_{1} = 1, …, q_{n} = 2k·q_{n−1}+q_{n−2}. Furthermore, p_{n}^{2}−q_{n}^{2}(k^{2}+1) = (−1)^{n}.

Then u_{1} = −q_{n} + \sqrt{q_{n}^{2}(k^{2} + 1) + (−1)^{n}}, r_{1} = q_{n} + \sqrt{q_{n}^{2}(k^{2} + 1) + (−1)^{n}}, and s_{1} = k ·q_{n} for each n ≥ 1. This implies that the matrix J_{1} which commutes with C_{1} will be one of the sequence of matrices determined by a triple (r_{1}, s_{1}, u_{1}) generated by values of q_{n}.

If q_{0} = 0 for n = 0, then r_{1} = 1, s_{1} = 0, u_{1} = 1, and J_{1} = I_{3} = C_{1}^{0}. For n = 1, q_{1} = 1, so that r_{1} = k + 1, s_{1} = k, and u_{1} = k − 1. The matrix J_{1} determined by these values is C_{1}. Now for n = 2, q_{1} = 2k, which makes r_{1} = 2k^{2}+2k+1, s_{1} = 2k^{2}, and u_{1} = 2k^{2} − 2k + 1. These values will produce J_{1} = C_{1}^{2}. It can be shown by math induction that C_{1}^{n} is generated by s = q_{n} · k, r = q_{n} + \sqrt{q_{n}^{2}(k^{2} + 1) + (−1)^{n}}, and u = −q_{n} + \sqrt{q_{n}^{2}(k^{2} + 1) + (−1)^{n}} so that the pattern will continue and clearly imply that J_{1} is a power of C_{1}.

\underline{Corollary 3}. Let J_{1} be a matrix of form J generated by r_{1}, s_{1}, u_{1}. J_{1} is a power of a matrix of form C if and only if

k = \frac{2s_{1}}{r_{1} − u_{1}}

is an integer. In this case, J_{1} is a power of the matrix of form C generated by the integer k.

As was seen in Theorem 4, two matrices J_{1} and J_{2} of form J commute if and only if

\frac{s_{1}}{r_{1} − u_{1}} = \frac{s_{2}}{r_{2} − u_{2}} .

If the ratio 2s/(r − u) for each of the matrices J_{1} and J_{2} equals an integer, then J_{1} and J_{2} are powers of a matrix of form C and obviously commute. The question remains, though, for what fractions p/q do there exist commuting matrices for which

\frac{2s}{r − u} = \frac{p}{q}.

Consider the rewritten form of the condition of Theorem 4

s_{2} = \frac{s_{1}(r_{2} − u_{2})}{r_{1} − u_{1}}.  (10)

The r, s, and u values in the table below for matrices J_{1} and J_{2} were computer– derived values to satisfy equation (10) and generate matrices that commute.

J_{1} matrix J_{2} matrix
r_{1}  s_{1}  u_{1}  \frac{2s_{1}}{r_{1}−u_{1}}  r_{2}  s_{2}  u_{2}  \frac{2s_{2}}{r_{2}−u_{2}}
5  2  1  \frac{1}{1}  12  5  2  \frac{1}{1}
17  4  1  \frac{1}{2}  72  17  4  \frac{1}{2}
37  6  1  \frac{1}{3}  228  37  6  \frac{1}{3}
65  8  1  \frac{1}{4}  528  65  8  \frac{1}{4}
101  10  1  \frac{1}{5}  1020 101  10  \frac{1}{5}

Notice that the fractions are of the form 1/n, and the values of r, s, and u seem to be forming a sequence for each 1/n. For each fraction p/q in lowest terms not equal to an integer, it is possible that there will exist a set of commuting matrices of form J such that for each J in the set,

\frac{2s}{r − u} = \frac{p}{q}.

It is conjectured, however, considering computer–generated examples similar to those listed above, that only fractions of the form 1/n admit a set of commuting matrices.

5. Monoids of Commuting Matrices. The earlier theorems established conditions for Symmetric Pythagorean Triple Preserving matrices to commute, thereby producing products that are symmetric. The goal was to find closed sets of SPTPM’s under matrix multiplication. From Theorem 5 and Corollary 3, it is seen that the value of the ratio 2s/(r − u) will generate different monoids. The identity matrix must be in each monoid, but since the values r = 1, s = 0, u = 1 produce the identity, the value of the ratio 2s/(r − u) for the identity is the indeterminate 0/0. The identity matrix, which is a SPTPM, will thus be considered an element in each monoid.

Let k be an integer. Let C_{k} denote the matrix of form C generated by the integer k. The set of all nonnegative powers of C_{k} (the identity taken as C_{k}^{0}) will form a closed set under matrix multiplication and thus be a monoid. Theorem 5 gives the method for completely describing each matrix in this monoid. Thus, for each integer k there exists a monoid of SPTPM’s such that the ratio 2s/(r − u) can be considered to equal k for all matrices in the monoid.

Next consider monoids corresponding to the fraction 1/2. Let

J_{0} = [ \begin{matrix}7 & 4 & 8 \\ 4 & 1 & 4 \\ 8 & 4 & 9 \end{matrix} ] ,

r = 4, s = 1, u = 0; and let

J_{1} = [ \begin{matrix}129 & 64 & 144 \\ 64 & 33 & 72 \\ 144 & 72 & 161 \end{matrix} ] ,

r = 17, s = 4, u = 1. If J_{2} is generated by r = 72, s = 17, u = 4, J_{3} by r = 305, s = 72, u = 17, J_{4} by r = 1292, s = 305, u = 72, and J_{5} by r = 5473, s = 1292, u = 305, then for each of these,

\frac{2s}{r − u} = \frac{1}{2}.

J_{1} = J_{0}^{2}, J_{2} = J_{0}^{3}, J_{3} = J_{0}^{4} = J_{1}^{2}, J_{4} = J_{0}^{5}, and J_{5} = J_{0}^{6} = J_{1}^{3} = J_{2}^{2}. The powers of J_{0} form a monoid, and since these powers will commute,

\frac{2s}{r − u} = \frac{1}{2}.

It can be shown that the powers of J_{0} will form the complete monoid for 1/2.

Notice for each J_{i} in the paragraph above, the s and u values of J_{i} are the r and s values, respectively, of J_{i−1}, for each J_{i}, r = 4s + u, or

\frac{2s}{r − u} = \frac{1}{2}.

The monoid for 1/2 seems to also be determined by the sequence 1, 0, 1, 4, 17, 72, 305, 1292, 5473, …, where the u, s, and r values (r = 4s + u) for a particular matrix in the monoid will be three consecutive elements of the sequence. (1, 0, 1 determines the identity, or J_{0}^{0}.) Again, it can be shown that the sequence listed above will generate all matrices in the monoid for 1/2.

In the general case, matrices in the monoid for the fraction 1/n will have u, s, and r values that come from the sequence 1, 0, 1, 2n, 4n^{2} + 1, 8n^{3} + 4n, 16n^{4} + 12n^{2} +1, …, where the elements of this sequence satisfy the recurrence relationship a_{j} = 2n · a_{j−1} + a_{j−2}. If u = a_{j−2}, s = a_{j−1}, and r = a_{j} = 2n · a_{j−1} + a_{j−2}, then it can be easily seen that

\frac{2s}{r − u} = \frac{1}{n}.

By a method similar to the proof of Theorem 5, it can be conclusively shown that the powers of an SPTPM J_{0}, with r = 2n, s = 1, and u = 0, will generate all matrices in the monoid for 1/n, and further, that the sequence listed above will generate all the matrices in the monoid.

6. Conclusion. The main focus of this paper was to find sets of Symmetric Pythagorean Triple Preserving Matrices (SPTPM) that are closed under matrix multiplication. The general form A of a Pythagorean Triple Preserving Matrix was first revised to develop a general form J for SPTPM’s with parameters r, s, and u. Developing closed sets of SPTPM’s involved finding products of symmetric matrices that would again be symmetric, but this implied the products of these symmetric matrices must be commutative. Since this is known to occur when the two matrices have a set of eigenvectors in common, the condition

\frac{s_{1}}{r_{1} − u_{1}} = \frac{s_{2}}{r_{2} − u_{2}} ,

that will guarantee when two SPTPM’s commute, was established by equating the eigenvectors of two SPTPM’s J_{1} and J_{2} generated by r_{1}, s_{1}, u_{1} and r_{2}, s_{2} u_{2}, respectively. From this condition then, it was determined that two types of monoids of commutative SPTPM’s are now known to exist, and each matrix in a particular monoid satisfies either the condition

\frac{2s}{r − u} − k,

k an integer, or

\frac{2s}{r − u} = \frac{1}{n},

n an integer.

\underline{References}

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

3. C. Vanden Eynden, Elementary Number Theory, Random House, New York, 1987.

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10. T. R. Lohmeier, Symmetric Pythagorean Triple Preserving Matrices, Masters Thesis, Southeast Missouri State University, Cape Girardeau, Missouri, 1996.

11. C. L. Sebaugh, Commutative Monoids of Symmetric Pythagorean Triple Preserving Matrices, Masters Thesis, Southeast Missouri State University, Cape Girardeau, Missouri, 1998.

James T. Bruening Department of Mathematics Southeast Missouri State University Cape Girardeau, MO 63701 email: jbruening@semovm.semo.edu

Tracy R. Lohmeier Cape Central High School 205 N. Caruthers Cape Girardeau, MO 63701

Cheryl L. Sebaugh Cape Central High School 205 N. Caruthers Cape Girardeau, MO 63701