primitive pythagorean triple, $\gcd(b,c)=1$ proof 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?
 A: After the point that $p\ |\ \frac{1}{2}(s^2-t^2)$ and $p\ |\ \frac{1}{2}(s^2+t^2)$ assume that, $\frac{1}{2}(s^2-t^2) = pk$ and $\frac{1}{2}(s^2+t^2) = pl$. Then, if you add two equations: $s^2 = p(k+l) \Rightarrow p\ |\ s^2 \Rightarrow p\ |\ s$. 
Do the same thing for $t$ and show that $p$ also divides $t$. Then if $p\ |\ s$ and $p\ |\ t$ it should also divide $a$ but in this case $(a, b, c)$ is not a PPT. (Contradiction.)
A: If $p|b$ and $p|c$ then it must be true that $p|a$ and therefore the triple is not primitive.
The equations are built to satisfy the pythagorean triple, $a^2 + b^2 = c^2$ (that I assume is the correct placement of the variables), and the gcd restrictions are there to keep scale multiples of the equation from consideration, those scale multiples would be the ones considered not primitive.
So other than checking the equations work, all that is left is to make sure the equation is primitive, which is to say not a multiple. I will leave the algebra for checking the equations, but as far as divisibility goes, the information I have given should give you good food for thought.
A: I think it is easier to prove instead that gcd $(a,b)=1$.
If $p|a$ then $p|st$ thus $p|s$ or $p|t$.
Since $p|b$ then $p|s^2-t^2=(s-t)(s+t)$. Thus $p|s \pm t$.
Now it is very easy to prove that ($p|s$ or $p|t$) and  $p|s \pm t$ implies that $p$ divides both $s$ and $t$. 
