Problem 1553

Proof. Let $x, y, z$ be a primitive Pythagorean triple. Lemma 11.2 tells us that $x$ is odd and $y$ is even, or vice versa. Since we have assumed that $y$ is even, $x$ and $z$ are both odd. Hence, $z+x$ and $z-x$ are both even, so that there are positive integers $r$ and $s$ with $r=(z+x) / 2$ and $s=(z-x) / 2$.

Since $x^{2}+y^{2}=z^{2}$, we have $y^{2}=z^{2}-x^{2}=(z+x)(z-x)$. Hence,
$$\left(\frac{y}{2}\right)^{2}=\left(\frac{z+x}{2}\right)\left(\frac{z-x}{2}\right)=r s$$

We note that $(r, s)=1$. To see this, let $(r, s)=d$. Since $d \mid r$ and $d \mid s$, $d \mid(r+s)=z$ and $d \mid(r-s)=x$. This means that $d \mid(x, z)=1$, so that $d=1$

Using Lemma 11.3, we see that there are integers $m$ and $n$ such that $r=m^{2}$ and $s=n^{2}$. Writing $x, y$, and $z$ in terms of $m$ and $n$, we have
$$\begin{array}{l} x=r-s=m^{2}-n^{2} \\ y=\sqrt{4 r s}=\sqrt{4 m^{2} n^{2}}=2 m n \end{array}$$
and
$$z=r+s=m^{2}+n^{2}$$

We see also that $(m, n)=1$, since any common divisor of $m$ and $n$ must also divide $x=m^{2}-n^{2}, y=2 m n$, and $z=m^{2}+n^{2}$, and we know that $(x, y, z)=1$. We also note that $m$ and $n$ cannot both be odd, for if they were, then $x, y$, and $z$ would all be even, contradicting the condition $(x, y, z)=1$. Since $(m, n)=1$ and $m$ and $n$ cannot both be odd, we see $m$ is even and $n$ is odd, or vice versa. This shows that every primitive Pythagorean triple has the appropriate form.

To see that every triple
$$\begin{array}{l} x=m^{2}-n^{2} \\ y=2 m n \\ z \equiv m^{2}+n^{2} \end{array}$$
where $m$ and $n$ are positive integers, $m>n,(m, n)=1$, and $m \not \equiv n(\bmod 2)$, forms a primitive Pythagorean triple, first note that
$$\begin{aligned} x^{2}+y^{2} & =\left(m^{2}-n^{2}\right)^{2}+(2 m n)^{2} \\ & =\left(m^{4}-2 m^{2} n^{2}+n^{4}\right)+4 m^{2} n^{2} \\ & =m^{4}+2 m^{2} n^{2}+n^{4} \\ & =\left(m^{2}+n^{2}\right)^{2} \\ & =z^{2} \end{aligned}$$

To see that these values of $x, y$, and $z$ are mutually relatively prime, assume that $(x, y, z)=d>1$. Then, there is a prime $p$ such that $p \mid(x, y, z)$. We note that $p \neq 2$, since $x$ is odd (because $x=m^{2}-n^{2}$ where $m^{2}$ and $n^{2}$ have opposite parity). Also, note that because $p \mid x$ and $p|z, p|(z+x)=2 m^{2}$ and $p \mid(z-x)=2 n^{2}$. Hence $p \mid m$ and $p \mid n$, contradicting the fact that $(m, n)=1$. Therefore, $(x, y, z)=1$, and $x, y, z$ is a primitive Pythagorean triple. This concludes the proof.