Solutions of Pythagorean Equation/Primitive/Proof 1

Proof
First we show that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is a Pythagorean triple:

So $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is indeed a Pythagorean triple.

Now we establish that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive:

to the contrary, that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is not primitive.

So there is a prime divisor $p$ of both $2 m n$ and $m^2 - n^2$.

That is:
 * $p \in \mathbb P: p \divides \paren {2 m n}, p \divides \paren {m^2 - n^2}$

Then from Prime Divides Power:
 * $p \divides \paren {2 m n}^2$ and $p \divides \paren {m^2 - n^2}^2$

Hence from Common Divisor Divides Integer Combination:
 * $p \divides \paren {m^2 + n^2}^2$

and from Prime Divides Power again:
 * $p \divides \paren {m^2 + n^2}$

So from Common Divisor Divides Integer Combination:
 * $p \divides \paren {m^2 + n^2} + \paren {m^2 - n^2} = 2 m^2$
 * $p \divides \paren {m^2 + n^2} - \paren {m^2 - n^2} = 2 n^2$

But $p \ne 2$ as, because $m$ and $n$ are of opposite parity, $m^2 - n^2$ must be odd.

So $p \divides n^2$ and $p \divides m^2$ and so from Prime Divides Power, $p \divides n$ and $p \divides m$.

But as we specified that $m \perp n$, this is a contradiction.

Therefore $\paren {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive.

Now we need to show that every primitive Pythagorean triple is of this form:

So, suppose that $\tuple {x, y, z}$ is any primitive Pythagorean triple given in canonical form.

From Parity of Smaller Elements of Primitive Pythagorean Triple, $x$ and $y$ are of opposite parity.

By definition of canonical form $x$ is even and $y$ and $z$ are both odd.

As $y$ and $z$ are both odd, their sum and difference are both even.

Hence we can define:
 * $s, t \in Z: s = \dfrac {z + y} 2, t = \dfrac {z - y} 2$.

Note that $s \perp t$ as any common divisor would also divide $s + t = z$ and $s - t = y$, and we know that $z \perp y$ from Elements of Primitive Pythagorean Triple are Pairwise Coprime.

Then from the Pythagorean equation:
 * $x^2 = z^2 - y^2 = \paren {z + y} \paren {z - y} = 4 s t$

Hence:
 * $\paren {\dfrac x 2}^2 = s t$

As $x$ is even, $\dfrac x 2$ is an integer and so $s t$ is a square.

So each of $s$ and $t$ must be square as they are coprime.

Now, we write $s = m^2$ and $t = n^2$ and substitute back:


 * $x^2 = 4 s t = 4 m^2 n^2$ and so $x = 2 m n$
 * $y = s - t = m^2 - n^2$
 * $z = m^2 + n^2$

Finally, note that:
 * $m \perp n$ from $s \perp t$ and Prime Divides Power
 * $m$ and $n$ have opposite parity otherwise $y$ and $z$ would be even.

Thus, our primitive Pythagorean triple is of the form $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$.