Solutions of Pythagorean Equation

Primitive Solutions of Pythagorean Equation
The set of all primitive Pythagorean triples is generated by:
 * $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$

where:
 * $m, n \in \Z$ are positive integers
 * $m \perp n$, that is, $m$ and $n$ are coprime
 * $m$ and $n$ are of opposite parity
 * $m > n$.

General Solutions of Pythagorean Equation
Let $x, y, z$ be a solution to the Pythagorean equation.

Then $x = k x', y = k y', z = k z'$, where:
 * $\left({x', y', z'}\right)$ is a primitive Pythagorean triple
 * $k \in \Z: k \ge 1$.

Proof 1
First we show that $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ is a Pythagorean triple:

So $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ is indeed a Pythagorean triple.

Now we establish that $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ is primitive:

Suppose to the contrary, that $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ is not primitive.

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

That is:
 * $p \in \mathbb P: p \mathop \backslash \left({2 m n}\right), p \mathop \backslash \left({m^2 - n^2}\right)$

Then from Prime Divides Power:
 * $p \mathop \backslash \left({2 m n}\right)^2$ and $p \mathop \backslash \left({m^2 - n^2}\right)^2$

Hence from Common Divisor Divides Integer Combination:
 * $p \mathop \backslash \left({m^2 + n^2}\right)^2$

and from Prime Divides Power again:
 * $p \mathop \backslash \left({m^2 + n^2}\right)$

So from Common Divisor Divides Integer Combination:
 * $p \mathop \backslash \left({m^2 + n^2}\right) + \left({m^2 - n^2}\right) = 2 m^2$
 * $p \mathop \backslash \left({m^2 + n^2}\right) - \left({m^2 - n^2}\right) = 2 n^2$

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

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

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

Therefore $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ is primitive.

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

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

From Parity of 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:
 * $x^2 = z^2 - y^2 = \left({z + y}\right) \left({z - y}\right) = 4 s t$

Hence:
 * $\left({\dfrac x 2}\right)^2 = s t$

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 $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$.

Proof 2
Let $\left({A,B,C}\right)$ be a Pythagorean Triple:


 * $A^2 + B^2 = C^2$

By the Pythagorean theorem, this equation describes the sides of a right triangle:

(picture goes here)

By the definitions of sine and cosine:

That is,

Next, we invoke Equiangular Triangles are Similar and Proportion is Equivalence Relation to write:

We construct the following restriction of $\tan \dfrac \theta 2$.

From Shape of Tangent Function, $\tan \dfrac \theta 2: \left({0 \,.\,.\, \pi}\right) \leftrightarrow \left({0 \,.\,.\, +\infty}\right)$ is a bijection.

Restrict $\tan \dfrac \theta 2$ on this interval so that its image is the set of strictly positive rational numbers:


 * $\tan \dfrac \theta 2: \tan^{-1}\left[{\Q_{>0}}\right]\cap \left({0 \,.\,.\, \pi}\right) \leftrightarrow \Q_{>0}$

Then $\displaystyle \tan \frac \theta 2 = \frac p q$ for any $\dfrac p q \in \Q_{>0}$, where $\dfrac p q$ is the canonical form of a rational number.

From the Double Angle Formulas:

Thus these proportions describe the sides of a right triangle:

(picture goes here)

By the Pythagorean theorem $\left({2pq,q^2 - p^2, q^2 + p^2}\right)$ is a Pythagorean triple.

That $p, q \in \Z_{>0}$ follows from $\dfrac p q \in \Q_{>0}$.

That $p \perp q$ follows from the assumption that $\dfrac p q$ was written in canonical form.

That every triple is of this form follows from the bijectivity of the tangent function as restricted above.

Recall $q^2 - p^2$ describes the side of a triangle, and so is positive.

Then $q^2 - p^2 > 0$ and therefore $q > p$.

It remains to be proven that:


 * this triple is primitive


 * $p$ and $q$ are of opposite parity.

General Solutions
Let $\left({x, y, z}\right)$ be non-primitive solution to the Pythagorean equation.

Let:
 * $\exists k \in \Z: k \ge 2, k \mathop \backslash x, k \mathop \backslash y$

such that $x \perp y$.

Then we can express $x$ and $y$ as $x = k x', y = k y'$.

Thus:
 * $z^2 = k^2 x'^2 + k^2 y'^2 = k^2 z'^2$

for some $z' \in \Z$.

Let:
 * $\exists k \in \Z: k \ge 2, k \mathop \backslash x, k \mathop \backslash z$

such that $x \perp z$

Then we can express $x$ and $z$ as $x = k x', z = k z'$.

Thus:
 * $y^2 = k^2 z'^2 - k^2 x'^2 = k^2 y'^2$

for some $y' \in \Z$.

Similarly for any common divisor of $y$ and $z$.

Thus any common divisor of any pair of $x, y, z$ has to be a common divisor of the other.

Hence any non-primitive solution to the Pythagorean equation is a constant multiple of some primitive solution.

Historical Note
This solution was known to Diophantus of Alexandria.