derivation of Pythagorean triples

For finding all positive solutions of the Diophantine equation

$x^2+y^2=z^2$(1)

we first can determine such triples  $x,y,z$  which are coprime.  When these are then multiplied by all positive integers, one obtains all positive solutions.

Let  $(x,y,z)$  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, $x$ and $y$ cannot both be even.  Neither can they both be odd, since because the square of any odd number is  $\equiv 1(mod4)$, the equation (1) would imply an impossible congruence  $2\equiv z^2(mod4)$.  Accordingly, one of the numbers, e.g. $x$, is even and the other, $y$, odd.

Write (1) to the form

$x^2=(z+y)(z-y).$(2)

Now, both factors (http://planetmath.org/Product) on the right hand side are even, whence one may denote

$z+y=:\mathrm{\hspace{0.33em}2}u,z-y=:\mathrm{\hspace{0.33em}2}v$(3)

giving

$z=u+v,y=u-v,$(4)

and thus (2) reads

$x^2=\mathrm{\hspace{0.33em}4}uv.$(5)

Because $z$ and $y$ are coprime and  $z>y>0$,  one can infer from (4) and (3) that also $u$ and $v$ must be coprime and  $u>v>0$.  Therefore, it follows from (5) that

$$u=m^2,v=n^2$$

where $m$ and $n$ are coprime and  $m>n>0$.  Thus, (5) and (4) yield

$x=\mathrm{\hspace{0.33em}2}mn,y=m^2-n^2,z=m^2+n^2.$(6)

Here, one of $m$ and $n$ is odd and the other even, since $y$ is odd.

By substituting the expressions (6) to the equation (1), one sees that it is satisfied by arbitrary values of $m$ and$n$.  If $m$ and $n$ have all the properties stated above, then $x,y,z$ are positive integers and, as one may deduce from two first of the equations (6), the numbers $x$ and $y$ and thus all three numbers are coprime.

Thus one has proved the

Theorem.  All coprime positive solutions  $x,y,z$,  and only them, are gotten when one substitutes for $m$ and $n$ to the formulae (6) all possible coprime value pairs, from which always one is odd and the other even and  $m>n$.