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
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 ,
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
is a perfect square.
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
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
Then we have
Case 1a, :
And since and are relatively prime positive integers of opposite parity so are and . Hence is a PPT.
Case 1b, :
Where is a PPT.
Case 2, is odd: So, there exists relatively prime positive integers of opposite parity, and , , such that
Note that since and are relatively prime then so are and .
Case 2a, 3 divides
set and Then choose and such that
Case 2b, 3 divides set and Then choose and such that
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
Solution: Since is primitive, (reduced to lowest terms) equals . (see Finding and ... ) So and . Then
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,