Solutions of Pythagorean Equation

Theorem

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;
• $m \perp n$, i.e. $m$ 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

Primitive Solutions

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({2 m n}\right)^2 + \left({m^2 - n^2}\right)^2\)	\(=\)	\(\displaystyle 4 m^2 n^2 + m^4 - 2 m^2 n^2 + n^4\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle m^4 + 2 m^2 n^2 + n^4\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({m^2 + n^2}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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, that $p \in \mathbb P: p \backslash \left({2 m n}\right), p \backslash \left({m^2 - n^2}\right)$.

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

Hence $p \backslash \left({m^2 + n^2}\right)^2$ from Common Divisor Divides Integer Combination, and from Prime Divides Power again, $p \backslash \left({m^2 + n^2}\right)$.

So:

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

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

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

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.

That is, $x$ 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 $\displaystyle s, t \in Z: s = \frac {z + y} 2, t = \frac {z - y} 2$.

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 All Elements of Primitive Pythagorean Triple are Coprime.

Then from the Pythagorean equation, we have:

$x^2 = z^2 - y^2 = \left({z+y}\right) \left({z-y}\right) = 4 s t$

Hence

$\displaystyle \left({\frac x 2}\right)^2 = s t$

As $x$ is even, $\displaystyle \frac x 2$ 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 $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$.

$\blacksquare$

General Solutions

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

• Let $\exists k \in \Z: k \ge 2, k \backslash x, k \backslash y$ 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 $\exists k \in \Z: k \ge 2, k \backslash x, k \backslash z$ 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.

$\blacksquare$

Historical Note

This solution was known to Diophantus of Alexandria.

Sources

• George F. Simmons: Calculus Gems (1992) $\text{A}.9$: Appendix