Primitive Pythagorean Numbers Are Coprime
Theorem
For three natural numbers $a,b,c$ satisfying $a^2 + b^2 = c^2$, if $\gcd (a,b,c) = 1$ then $$ \gcd (a,b) = 1 \\ \gcd (b,c) = 1 \\ \gcd (c,a) = 1 $$
Explanation
At first glance this seems obvious for Pythagorean numbers or whatever, but if you think carefully about what a common divisor is, it is not so obvious. For example, without the condition of being Pythagorean numbers, $\gcd (6,10,15) = 1$ holds, yet each pair of the numbers has its own common divisor. Strategy: The proof fundamentally presupposes the following two lemmas.
Another Representation of Pythagorean Numbers: For three natural numbers $a,b,c$ satisfying $a^2 + b^2 = c^2$, $$ \begin{align*} a =& st \\ b =& {{s^2 - t^2 } \over {2}} \\ c =& {{s^2 + t^2 } \over {2}} \end{align*} $$ there exist two coprime odd numbers $s>t$ satisfying these.
In the process of proving the above theorem, we can obtain that $\gcd (s,t) = 1$ when $(a,b,c)$ is a primitive Pythagorean triple.
A prime dividing a number divides at least one of its factors: For $n : = d_{1} d_{2} \cdots d_{r}$, if $p \mid n$ then $p$ must divide one of $d_{1} , d_{2} , \cdots , d_{r}$.
Proof
Part 1. $\gcd (a,b) = 1$
Assume that a prime $g$ is a common divisor of $$ a = st \\ \displaystyle b = {{s^2 - t^2 } \over {2}} $$
Then, since $g \mid st$, we must have $g \mid s$ or $g \mid t$, and since $\displaystyle g \mid {{s^2 - t^2 } \over {2}}$, we must have $g \mid (s-t)$ or $\displaystyle g \mid {{s + t } \over {2}}$. Here, if $g \mid s$, then from the assumption of being a primitive Pythagorean number we have $\gcd (s,t)=1$, so $g \nmid t$. Therefore $$ g \nmid (s-t) \\ g \nmid {{s + t } \over {2}} $$ which is a contradiction.
Part 2. $\gcd (b,c) = 1$
Assume that a prime $h$ is a common divisor of $$ b = {{s^2 - t^2 } \over {2}} \\ c = {{s^2 + t^2 } \over {2}} $$
Then, since $$ h \mid {{s^2 - t^2 } \over {2}} \\ h \mid {{s^2 + t^2 } \over {2}} $$ we have $$ h \mid \left( {{s^2 - t^2 } \over {2}} - {{s^2 + t^2 } \over {2}} \right) \\ h \mid \left( {{s^2 - t^2 } \over {2}} + {{s^2 + t^2 } \over {2}} \right) $$ Rearranging, $$ h \mid s^2 \\ h \mid t^2 $$ but then $\gcd (s,t) \ne 1$, which is a contradiction.
Part 3. $\gcd (c,a) = 1$
Assume that a prime $k$ is a common divisor of $$ a = st \\ c = {{s^2 + t^2 } \over {2}} $$
Then, since $k \mid st$, we must have $k \mid s$ or $k \mid t$, and $\displaystyle k \mid {{s^2 + t^2 } \over {2}}$. Here, if $k \mid s$, then from the assumption of being a primitive Pythagorean number we have $\gcd (s,t)=1$, so $k \nmid t$. Therefore $\displaystyle k \nmid {{s^2 + t^2 } \over {2}}$, which is a contradiction.
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