derivation of Pythagorean triples
Let be a solution of the mentioned kind. Then the numbers are pairwise coprime, since by (1), a common divisor of two of them is also a common divisor of the third. Especially, and cannot both be even. Neither can they both be odd, since because the square of any odd number is , the equation (1) would imply an impossible congruence . Accordingly, one of the numbers, e.g. , is even and the other, , odd.
Write (1) to the form
Now, both factors (http://planetmath.org/Product) on the right hand side are even, whence one may denote
and thus (2) reads
Because and are coprime and , one can infer from (4) and (3) that also and must be coprime and . Therefore, it follows from (5) that
where and are coprime and . Thus, (5) and (4) yield
Here, one of and is odd and the other even, since is odd.
By substituting the expressions (6) to the equation (1), one sees that it is satisfied by arbitrary values of and
. If and have all the properties stated above, then are positive integers and, as one may deduce from two first of the equations (6), the numbers and and thus all three numbers are coprime.
Thus one has proved the
Theorem. All coprime positive solutions , and only them, are gotten when one substitutes for and to the formulae (6) all possible coprime value pairs, from which always one is odd and the other even and .
- 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).
|Title||derivation of Pythagorean triples|
|Date of creation||2013-03-22 18:34:40|
|Last modified on||2013-03-22 18:34:40|
|Last modified by||pahio (2872)|