Let $s, t \in \mathbb{Z}^+$, be both odd, $s > t \geq 1$, $\gcd(s, t) = 1$. If we set
$a = st,\quad b = \frac{1}{2}(s^2 - t^2),\quad c = \frac{1}{2}(s^2 + t^2)$
then $(a, b, c)$ is primitive pythagorean triple.
I tried to prove it but I am not sure of my method. Could you please check my solution if there exists something senseless.
What I tried:
Suppose there exists a $p$ which is prime number, such that $p\ |\ b$ and $p\ |\ c$.
$p\ |\ b = \frac{1}{2}(s^2 - t^2) \Rightarrow p\ |\ 2b = s^2 - t^2 \Rightarrow p\ |\ s^2\ \textrm{and}\ p\ |\ t^2 \Rightarrow p\ |\ s\ \textrm{and}\ p\ |\ t\\ p\ |\ c = \frac{1}{2}(s^2 + t^2) \Rightarrow p\ |\ 2c = s^2 + t^2 \Rightarrow p\ |\ s^2\ \textrm{and}\ p\ |\ t^2 \Rightarrow p\ |\ s\ \textrm{and}\ p\ |\ t$
But we know that $\gcd(s, t) = 1$ and if $p\ |\ s$ and $p\ |\ t$, then this is contradiction.
Is this true?