Proving that Pythagorean triples are relatively prime if s and t are odd

Question

Problem: Let s and t be odd integers with $ s > t \geq 1$ and $\gcd(s,t)=1$. Prove that the three numbers

$ st , \frac{s^2-t^2}{2}, $ and $\frac{s^2+t^2}{2} $

are pairwise relatively prime.

Proof: Let $s=2k_1+1$ and $t=2k_2+1$, where $\gcd(k_1,k_2)=1$.

$st=(2k_1+1)(2k_2+1)=4k_1k_2+2k_1+2k_2+1=2(2k_1k_2+k_1+k_2)+1$. So st is odd.

$\frac{s^2-t^2}{2}=\frac{(2k_1+1)^2-(2k_2+1)^2}{2}=\frac{4k_1^2+4k_1+1-4k_2^2-4k_2-1}{2}=2(k_1^2+k_1-k_2^2-k_2)$. So $\frac{s^2-t^2}{2}$ is even.

$\frac{s^2+t^2}{2}=\frac{(2k_1+1)^2+(2k_2+1)^2}{2}=\frac{4k_1^2+4k_1+1+4k_2^2+4k_2+1}{2}=2(k_1^2+k_1+k_2^2+k_2)+1$. So $\frac{s^2+t^2}{2}$ is odd.

So, $\gcd(st,\frac{s^2-t^2}{2})=1$, because st is odd and $\frac{s^2-t^2}{2}$ is even. So, $\gcd(\frac{s^2+t^2}{2},\frac{s^2-t^2}{2})=1$, because $\frac{s^2-t^2}{2}$ is odd and $\frac{s^2-t^2}{2}$ is even. But st and $\frac{s^2+t^2}{2}$ are both odd, $\dots$

This is where I am stuck can is it correct for me to choose $k_1$ and $k_2$ to be relatively prime to each other? If not can I use that fact to show that the $\gcd(st,\frac{s^2+t^2}{2})=1$?

Answer 1

If a prime $p\ne2$ divides into $\text{gcd}(st,\frac{s^2+t^2}2)$, then either $p$ divides $s$, or it divides $t$. It cannot divide into both of them, because $\text{gcd}(s,t) = 1$. Therefore $p$ cannot divide into $s^2+t^2$, and we arrive at a contradiction.

The case $p=2$ is slightly different. You know $st$ is odd. Show that only one of $\frac{s^2+t^2}2$ and $\frac{s^2-t^2}2$ can be even.

Answer 2

Note of $p$ divides any two of them, it should divide the third also. So just concentrate on the last two: $$ p |\frac{s^2-t^2}{2}, ~\hbox{and}~p |\frac{s^2+t^2}{2} \Rightarrow p | s^2, ~\hbox{and}~~ p |t^2$$ since if $p$ divides two numbers, it should divide their sum and difference.

But $s$ and $t$ are co-prime, so no such $p$ exists.

Answer 3

Suppose prime $\,p\,$ divides two of $\,(a,b,c) = (st,(s^2-t^2)/2,(s^2+t^2)/2).\,$ Then it divides the third too, by $\,a^2\!+b^2\!-c^2=0.\,$ By Bezout $\,(s,t)=1\, \Rightarrow\, \color{#c00}{is+jt=1}\,$ for some $\,i,j\in \Bbb Z.\,$ Hence

$$p\mid \color{#0a0}{a,b,c}\, \Rightarrow\, p\mid i^2(\color{#0a0}{b\!+\!c})+2ij\color{#0a0}a+j^2(\color{#0a0}{c\!-\!b})= i^2s^2\!+2ijst+j^2t^2 = (\color{#c00}{is\!+\!jt})^2 = \color{#c00}1\ \Rightarrow\Leftarrow$$

Remark $\ $ Or, by ideals $\,(p)\supseteq(a,b,c)\,\Rightarrow\, (p)\supseteq (b\!+\!c,a,c\!-\!b)=(s^2,st,t^2)=(s,t)^2 = (1)$