proof of Pythagorean triplet
For to form a Pythagorean triple, each of must be squares. So where are coprime, . So we have
and are a Pythagorean triple. But this needn’t be primitive: If are odd, then , so not all of are relatively prime.
Suppose are pairwise coprime. Then , and it follows that . Thus , i.e. is odd. Furthermore, implies . And since the sum/difference of two integers is odd iff one is even, and the other is odd, only one of is odd. Thus, . Conversely, if are coprime, and exactly one of is odd, then ; thus, , are coprime. From the fact that if have opposite parity and it follows that , are also coprime. And since and it follows that are coprime. So the conditions the Pythagorean triple is primitive, and are coprime and exactly one of them is odd are equivalent.
So if satisfy and are pairwise coprime, then is odd, and exactly one of are odd and the other is even.
Let be coprime positive integers of opposite parity, . Set
since , . Clearly, .
Now we prove that any primitive Pythagorean triple can be generated choosing odd coprime integers.
Let be odd coprime integers, . Let . Then .
Since , the statement follows from the fact that have opposite parity since in this case . Since , is odd, and since and is odd, have opposite parity. ∎
Substituting , in yields that is a primitive Pythagorean triple.
To see that any primitive Pythagorean triple is of this form:
Let be positive coprime integers satisfying . Then have opposite parity, and is odd. Furthermore .
Suppose is odd. Since is a square, and and it follows that are coprime and consequently each of them is square. This gives where are odd coprime integers, and we get
Now let be a square. Without loss of generality we can set , where are odd coprime integers. So we have , and since and are coprime, each of them must itself be a square. So we have
where the right-hand side numbers are biquadratic integers. So the question if the area of a right triangle with integer sides is square is equivalent to asking if has a solution in positive integers.
|Title||proof of Pythagorean triplet|
|Date of creation||2013-03-22 14:06:52|
|Last modified on||2013-03-22 14:06:52|
|Owner||Thomas Heye (1234)|
|Last modified by||Thomas Heye (1234)|
|Author||Thomas Heye (1234)|