📂Number Theory

Primitive Pythagorean triples are coprime.

Theorem

Given three natural numbers satisfying $a^2 + b^2 = c^2$, if $a,b,c$ then $$\gcd (a,b) = 1 \\ \gcd (b,c) = 1 \\ \gcd (c,a) = 1$$

Explanation

At first glance, it might seem obvious whether it’s about Pythagorean triples or anything else, but it’s not necessarily the case when considering the greatest common divisor. For example, without the condition of being Pythagorean triples, $\gcd (6,10,15) = 1$ would be true, but each pair of numbers would have their own common divisor. Strategy: The proof essentially presupposes the following two lemmas.

Another expression of Pythagorean triples: For three natural numbers satisfying $a^2 + b^2 = c^2$, if $a,b,c$, then $$\begin{align*} a =& st \\ b =& {{s^2 - t^2 } \over {2}} \\ c =& {{s^2 + t^2 } \over {2}} \end{align*}$$ there exist two coprime odd numbers $s>t$.

In the process of proving the above theorem, it can be derived that when $(a,b,c)$ is a primitive Pythagorean triple, then $\gcd (s,t) = 1$.

A prime that divides a number also 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. $\gcd (a,b) = 1$

Let’s assume that the prime $g$ is $$a = st \\ \displaystyle b = {{s^2 - t^2 } \over {2}}$$ a common factor.

Then, since $g \mid st$, it must be either $g \mid s$ or $g \mid t$, and since $\displaystyle g \mid {{s^2 - t^2 } \over {2}}$, it must be either $g \mid (s-t)$ or $\displaystyle g \mid {{s + t } \over {2}}$. Here, if we suppose $g \mid s$, given the assumption of it being a primitive Pythagorean number, $\gcd (s,t)=1$ leads to $g \nmid t$. Therefore, $$g \nmid (s-t) \\ g \nmid {{s + t } \over {2}}$$ which is a contradiction.

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Part 2. $\gcd (b,c) = 1$

Let’s assume that the prime $h$ is $$b = {{s^2 - t^2 } \over {2}} \\ c = {{s^2 + t^2 } \over {2}}$$ a common factor.

Then, $$h \mid {{s^2 - t^2 } \over {2}} \\ h \mid {{s^2 + t^2 } \over {2}}$$ leads to $$h \mid \left( {{s^2 - t^2 } \over {2}} - {{s^2 + t^2 } \over {2}} \right) \\ h \mid \left( {{s^2 - t^2 } \over {2}} + {{s^2 + t^2 } \over {2}} \right)$$. Summarizing, $$h \mid s^2 \\ h \mid t^2$$, this means $\gcd (s,t) \ne 1$, which leads to a contradiction.

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Part 3. $\gcd (c,a) = 1$

Let’s assume that the prime $k$ is $$a = st \\ c = {{s^2 + t^2 } \over {2}}$$ a common factor.

Then, since $k \mid st$, it must be either $k \mid s$ or $k \mid t$, and $\displaystyle k \mid {{s^2 + t^2 } \over {2}}$. Here, if we suppose $k \mid s$, given the assumption of it being a primitive Pythagorean number, $\gcd (s,t)=1$ leads to $k \nmid t$. Therefore, $\displaystyle k \nmid {{s^2 + t^2 } \over {2}}$, which is a contradiction.

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