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

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

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

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

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

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

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.

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

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.

**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.

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**

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,