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# Multiplying Pythagorean Triples

This section in pdf form. triples11.pdf

Usually when people speak of multiplying Pythagorean triples they are referring to multiplying the hypotenuses of of two triples to generate another Pythagorean triple. For example, if and are Pythagorean triples then is also a Pythagorean triple. However, Pythagorean triples can be generated by multiplying the other sides of two triples as well. That is, not only does generate a Pythagorean triple, so does and

Preliminaries

We will first derive identities for multiplying the sums of two squares and for multiplying the differences of two squares to give the sum of two squares and the difference of two squares respectively.

Let and be elements of an integral domain. Then,

Hence, we have

 (24)

Set and substitute into equation (24) to get Fibonacci's Identity for multiplying the sums of two squares,

 (25)

Set , to get the Diophantus identity for multiplying the differences of two squares,

 (26)

Each of these identities is a special case of Brahmagupta's identity,

 (27)

Set and in (27) to get equations (25) and (26) respectively.

Multiplying Pythagorean triples

Note that , , and are equivalent methods of writing a Pythagorean triangle.

If and are two Pythagorean triples then, from equations(25) and (26), we have

(a)

(b)

(c)

(d)

(e)

From which we will define the following multiplications where and are Pythagorean triples. And the over bars designate multiplier and multiplicand.

If and then, from items ,

i.

ii.

iii.

iv.

v.

Examples

The ordered triples and are Pythagorean triples, hence from , we get the resultant Pythagorean triples,

Let be odd. Then clearly, from items i through v, if is a primitive Pythagorean triple (PPT) then so are

Example: Since is a PPT, so are

and

Supplemental

Claim 5   Let and be vectors in 3-space. Then, if and are also primitive Pythagorean triples, where and are of opposite parity, the scalar product

is a perfect square.

Proof. Without loss of generality let be odd and even. So, since both vectors are also PPTs, there exists positive integers and such that

and

and

Where and are relative prime, have opposite parity, and for and 2.

Hence we have,

Similarly it can be shown that the sums , , and are also perfect squares.

Example:    Let be a PPT where a is even, then each of the sums and is a perfect square.

Theorem 5   Let be greater than Then is a primitive Pythagorean triple (PPT) if and only if there exists another PPT, , such that ( equals either or

Proof. We know that since is a PPT then and have opposite parity. That is, one of and is odd and the other is even. We also know that 3 divides exactly one of and . Label such that 3 divides . Then there are two cases: is even, or is odd.

Case 1, is even: So, there exists relatively prime positive integers of opposite parity, and , , such that

and

where 3 divides exactly one of and .

If , set equal to the greater of and and equal to the lesser. Then choose and such that

and

Then we have

Case 1a, :

Therefore

And since and are relatively prime positive integers of opposite parity so are and . Hence is a PPT.

Case 1b, :

Therefore

Where is a PPT.

Case 2, is odd: So, there exists relatively prime positive integers of opposite parity, and , , such that

and

Note that since and are relatively prime then so are and .

Case 2a, 3 divides

set and Then choose and such that

and

We have,

Therefore, if , . And is a PPT. And if then and .

Case 2b, 3 divides set and Then choose and such that

and

We have,

Therefore . And is a PPT.

And going in the other direction, if is a PPT then either or is a PPT.

Proof.     If and or both not primitive then implies . And . Since , can not divide since . Hence or .

Similarly, if then or . That is, and . This implies that and , a contradiction. Hence either or is primitive.

That is, if divides then is a primitive Pythagorean triple if and only if there exists a primitive Pythagorean triple such that

Examples

and are primitive Pythagorean triples where 3 divides the even side.

Problem 1: Find a PPT such that

and

Solution: Since is primitive, (reduced to lowest terms) equals . (see Finding and ... ) So and . Then

And

Similarly, since is primitive, So and . Since , we have

is a PPT where 3 divides the odd side.

Problem 2: Find a PPT such that

Solution: Since is a PPT, And 3 divides , therefore and So,

Next: Finding parametric equations for Up: Pythagorean Triples, etc. Previous: Primitive Pythagorean triangles where   Contents
f. barnes 2008-04-29
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