A question on the Dickson's method of generating Pythagorean triples. I was trying to experiment with the Dickson's method of generating triples. I found out the values of $r, s,$ and $t$ and based on statistical evidence, found that $s \ge r$. Can we prove this generally for all triples?
My condition is that $s > t$. This implies $r + t < r + s$
Thanks,
S Sandeep
 A: You have
$$r^2=2st\qquad s>t$$
If $s$ and $t$ have a common factor, then $r$ would contain that factor as well. You could simply cancel that factor from all variables without affecting one of the relevant equalities or inequalities. So w.l.o.g. you may assume that $s$ and $t$ are coprime. In that case, exactly one of the two must be even. Furthermore, since $r^2$ is a square number, it must contain every prime factor an even number of times. So except for the fator of $2$, every prime factor will occur an even number of times in either $s$ or $t$, since it can't occur in both coptime numbers.
Case 1: $s$ is even.
Then there exist integers $u,v$ such that $s=2u^2$, $t=v^2$ and $r=2uv$. Now $s\ge r$ if and only if $u\ge v$. So let's assume the contrary, $u<v$. One example would be
$$u=4\quad v=5\quad s=32\quad t=25\quad r=40\quad r^2=2st=1600$$
So this satisfies $s>t$ but violates $s\ge r$, thereby disproving your claim!
Case 2: $t$ is even.
Since we already have a counterexample, there is no need to investigate this case as well.
