proof of Pythagorean triples
If , and are positive integers such that
then is a Pythagorean triple. If , and are relatively prime in pairs then is a primitive Pythagorean triple. Clearly, if divides any two of , and it divides all three. And if then . That is, for a positive integer , if is a Pythagorean triple then so is . Hence, to find all Pythagorean triples, it’s sufficient to find all primitive Pythagorean triples.
Let , and be relatively prime positive integers such that . Set
which, after cancelling and rearranging terms, becomes
There are two cases, either and are of opposite parity, or
they or both odd. Since , they can not both be
Case 1. and of opposite parity, i.e., . So 2 divides b since is odd. From equation (2), divides . Since then , therefore also divides . And since , divides . Therefore . Then
Case 2. and both odd, i.e., . So 2 divides . Then by the same process as in the first case we have
Then , and . Substituting those values for and into (5) we get
where , , and and are of opposite parity. Therefore (6), with a and b interchanged, is identical to (4). Thus since , as in (4), is a primitive Pythagorean triple, we can say that is a primitive pythagorean triple if and only if there exists relatively prime, positive integers and , , such that .
|Title||proof of Pythagorean triples|
|Date of creation||2013-03-22 14:28:05|
|Last modified on||2013-03-22 14:28:05|
|Last modified by||fredlb (5992)|