The Trinary Tree

7/27/2019 The Trinary Tree

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The Trinary Tree(s) underlying Primitive Pythagorean

Triples

H. Andres Lnnemo

June 8 2000

==========

FOUNDATION

==========

I've known this a long time:

Let ( a, b, c ) be a Pythagorean Triple

a2 + b2 = c2 a,b,c > 0

Parameterize a,b,c as follows

a = q + m m,n,q > 0

b = q + n

c = q + m + n

Substitute

(q + m)2 + (q + n)2 = (q + m + n)2

Expand, cancel common terms, and take square root to get

q = sqr(2*m*n)

Thus, Pythagorean Triples can be generated by finding m,nsuch that 2*m*n is a perfect square.

The following observations are offered without proof:

* If m,n are relatively prime then the Pythagorean

Triple will also be relatively prime. This is

known as a Primitive Pythagorean Triple, or PPT.

* q will always be even.

* In PPTs, either m or n must be even and the other odd

This is also true for a and b.

* q is larger than m or n or both

* q is smaller than m + n

The reverse equations are easily derived:

m = c - b

n = c - a

q = a + b - c
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========================

LOOK, THEY COME IN PAIRS

========================

This is the new twist:

Traditionally, q is assumed to be the positive root. However

if you instead choose the negative root, a different

Pythagorean

Triple is formed with a or b or both being negative. Discard

the

sign(s) and you have a perfectly good Pythagorean Triple.

Thus each valid combination of m and n produce two Triples.

Again,

if m and n are relatively prime so will be a, b and c in both

Triples.

One of the triples will have all positive terms, and in theother

a or b will be negative. The c's will always have the same

sign,

assumed positive without loss of generality.

The odd/even pattern in PPTs will be identical.

The Signed Primitive Pythagorean Triple will be known as the

SPPT.

For example:

let m = 25 and n = 8

q = sqr( 2 * 25 * 8 ) = 20

a = q + m = 45 a' = -q + m = 5

b = q + n = 28 b' = -q + n = -12

c = q + m + n = 53 c' = -q + m + n = 13

The PPT is (45,28,53) and the SPPT is (5,-12,13) which

corresponds to a PPT of (5,12,13).

The c' value of the SPPT will always be smaller than the c

value

of the PPT.

=================

TREE CONSTRUCTION

=================

The construction of the trinary tree is based on the

observation

that PPTs and SPPTs always come in pairs and every valid PPT

can

generate three, and only three, SPPTs simply by changing the


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signs

on a and b.

Start with (3,4,5) as the root node called "P".

Build a table of values:

A B C M N Q Q' A' B' C'

=== === === === === === === === === ===

PPT 3 4 5 1 2 2 -2 -1 0 1

(later)

SPPT -3 4 5 1 8 -4 4 5 12 13

PPT

SPPT 3 -4 5 9 2 -6 6 15 8 17

PPT

SPPT -3 -4 5 9 8 -12 12 21 20 29

PPT

You can now read the three child nodes from the table:

They are (5,12,13), (15,8,17) and (21,20,29) called P1, P2 andP3

respectively.

Repeat the process for each child node to build as large a

tree

as you want.

For example:

A B C M N Q Q' A' B' C'

=== === === === === === === === === ===

PPT 5 12 13 1 8 4 -4 -3 4 5

SPPT

SPPT -5 12 13 1 18 -6 6 7 24 25

PPT

SPPT 5 -12 13 25 8 -20 20 45 28 53

PPT

SPPT -5 -12 13 25 18 -30 30 55 48 73

PPT

Notice that the first row points back to the parent PPT when

the sign is removed from the PPT. Also the sign pattern will

tell which branch was taken.

As you build the tree you will notice that the c's are always

increasing as you traverse from the root, that is the list of

PPTs is 'quasi-sorted' in a heap sort sense.

==================================

THE SELF ROOTED NATURE OF THE TREE

==================================

Something funny happens at the apparent parent of (3,4,5)

which is (1,0,1).


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(a2 b2 c2) = (a b c)( -1 -2 -2 ) = (a b c)*T2

( 2 1 2 )

( 2 2 3 )

(a3 b3 c3) = (a b c)( 1 2 2 ) = (a b c)*T3

( 2 1 2 )

( 2 2 3 )

(a0 b0 c0) = (a b c)( -1 -2 -2 ) = (a b c)*T0

( -2 -1 -2 )

( 2 2 3 )

And this is a remarkable result. It means that any

Pythagorean

Triple can generate three new Triples by means of matrix

multiplications with T1, T2 and T3 with larger c's, and can

generate a signed version of a smaller c triple with a matrix

multiplication with T0. If the triple is a PPT the child

nodes will be PPTs and the T0 transform will yield a SPPT

corresponding to the parent.

=================

SPANNING THE PPTS

=================

The trinary tree covers the entire set of PPTs completely and

uniquely. The unique part is inherent from the construction of

the tree. Each node has only one unique SPPT 'mnq-twin' and

thus has only one parent.

The completely part requires a proof by contradiction.

Suppose you

have a PPT which is not spanned by the tree. It would still

have a

'mnq-twin' SPPT with a smaller c. And the absolute values of

those

numbers would form another PPT.

This PPT can't be on the tree either or the original one would

be,

and so on. Since the c's are decreasing, and are bounded by

zero,

this sequence must terminate in a different 'self-parent' than

(1,0,1) or (0,1,1). Since everything is relatively prime

these two

are the only two triples where q = 0, since q = 0 implies m =0 or

n = 0 (Remember q = sqr(2*m*n)).

================================

MNQ-TWINS IN LINEAR ALGEBRA FORM

================================


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T

------------------------------------>

(a b c) -----> (m n q) -----> (m n -q) -----> (a' b' c')

InvP U4 P

InvP = ( 0 -1 1 ) U4 = ( 1 0 0 ) P = ( 1 0 1 )

( -1 0 1 ) ( 0 1 0 ) ( 0 1 1 )

( 1 1 -1 ) ( 0 0 -1 ) ( 1 1 1 )

U4 * U4 = I

T = InvP * U4 * P = ( -1 -2 -2 )

( -2 -1 -2 )

( 2 2 3 )

Note, T * T = I

Proof: T * T = InvP * U4 * P * InvP * U4 * P = InvP * U4 * U4

* P

= InvP * P = I

Thus the T transform will find the 'mnq-twin' of any triple.

========================================

TREE CONSTRUCTION IN LINEAR ALGEBRA FORM

========================================

Parent -------------------------> Child

Signed Variation MNQ-Twin

PPT SPPT PPT

(a b c) -----> ( a' b' c ) -----> ( a" b" c" )

Ui T i = 1,2,3

U1 = ( -1 0 0 ) U2 = ( 1 0 0 ) U3 = ( -1 0 0 )

( 0 1 0 ) ( 0 -1 0 ) ( 0 -1 0 )

( 0 0 1 ) ( 0 0 1 ) ( 0 0 1 )

T1 = U1 * T = ( 1 2 2 )( -2 -1 -2 )

( 2 2 3 )

T2 and T3 multiply out identically to the results in the

previous

section.

================================

EIGENVECTORS AND SELF ROOTEDNESS


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================================

The Eigenvectors of T are also interesting.

The characteristic equation of T is

-x3 + x2 + x - 1 = 0

Which yields roots of 1, 1, and -1. Since there is a

double root at one the eigenvectors require two parameters.

A set of corresponding eigenvectors are:

( s1, s2, s1+s2 ) and ( s3, s3, s3 )

Since (1,1,1) can never be a Pythagorean triple when

multiplied

by any length (except the trivial zero), it can be discarded.

Applying the Pythagorean constraint to the first family of

vectors:

(s1)2 + (s2)2 = (s1 + s2)2

s12 + s22 = s12 + 2*s1*s2 + s22

0 = 2*s1*s2

Therefore s1 = 0 or s2 = 0 or both. Both is again the trivial

solution and can be discarded. That leaves two vectors:

( s1, 0, s1 ) and ( 0, s2, s2 ) which reduce to ( 1, 0, 1 )

and ( 0, 1, 1 ) when relative primeness is introduced.

These are precisely the two self rooted triplets that lead to

all the odd/even/odd and even/odd/odd PPTs respectively.

====================

AN INTERESTING TWIST

====================

Reversing the first two rows of T has the effect of swapping

the odd/even pattern of a and b. This has the effect of

making (1 0 1) the parent of (0 1 1) and vice versa. So

instead

of two separate self rooted trees, you get sort of 'Siamese'

trees joined at the head where the tiers alternate between

even/oddand odd/even triples.

The eigenvalues of this matrix are -1, -1, and 1 with

eigenvectors:

( s1, s2, (s1+s2)/2 ) and (1,1,2)

Applying the Pythagorean constraint to these shows no real

solutions possible. Therefore there can be no other


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self-rooted trees based on this matrix.

The Trinary Tree

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