Primitive Pythagorean triples with the help of trigonometry

 Not a puzzle, just a quick note.

Primitive pythagorean triples are of the form $m^2 - n^2, 2mn, m^2 + n^2$ with $m,n$ relatively prime and of opposite parities.

We will use trigonometry to help derive this.

Let $a,b,c$ with $a^2 + b^2 = c^2$ be a primitive pythagorean triplet. This means $\gcd(a,b,c) = 1$. Note that this also implies that $a,b,c$ are pairwise coprime. Thus exactly one of $a,b,c$ is even. It cannot be $c$ as that would imply $4$ divides $a^2 + b^2$ for $a,b$ odd (which is not possible). 

Thus assume $b$ is even. 

We can now prove that 

$\textbf{Proposition:}$ The highest power of $2$ that divides $b$ is different from the highest power of $2$ that divides $a+c$. 

$\textbf{Proof:}$ Since $a,c$ are odd and relatively prime, we must have that $\gcd(a+c, c-a) = 2$. Thus $a+c = 2u$ and $c-a = 2v$ for some odd $u,v$. Now if $b = 2w$ for some odd $w$ (if we assume that the highest power of $2$ that divides $b$ is same as $a+c$), then $b^2 = (c+a)(c-a) \implies w^2 = uv$. Since $w^2 = 1 \mod 4$, this implies $u = v$ mod $4$, and thus $a = u - v = 0 \mod 4$, not possible.

Now consider the triangle $$B = (0,0), C = (a, 0), A = (a,b)$$

Draw the anglular bisector $BD$ of angle $B$ such that $D$ lies on $AC$. By the angular bisector theorem, $CD/DA = a/c$ and hence $CD = \frac{ab}{a+c}$. Thus we have that

$$\tan \left(\frac{B}{2}\right) = \frac{CD}{BC} = \frac{b}{a+c} = \frac{n}{m}$$

For some $n, m$ relatively prime.  Since the highest power of $2$ that divides $b$ is different from that of $a+c$, $n,m$ are of opposite parities.

Now using $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$ we get

$$\frac{b}{a} = \tan B = \frac{2 n/m}{1 - (n/m)^2} = \frac{2nm}{m^2 - n^2}$$

We can show that $2mn$ and $m^2 -n^2$ are relatively prime if $m,n$ are relatively prime and of opposite parities: if a prime $p > 2$ divides the numerator, it has to divide one of $m,n$ and cannot divide the denominator also. The denominator is also odd, so their $\gcd$ cannot be $2$.

Thus $$b = 2mn, a = m^2 - n^2, c = m^2 + n^2$$