I found out that in a primitive pythagorean triple $$a^2+b^2=c^2$$ the difference $d=|a-b|$ (which must be odd) can occur, if and only if we can write $$d=a^2-2b^2$$ with positive coprime integers $a,b$. Moreover, $d$ is a possible difference if and only if $-d$ is a possible difference. We can replace the pair $(a/b)$ by $(a+2b/a+b)$ to get a solution of the desired form.
When can an odd integer $d$ be written as $d=a^2-2b^2$ with positive coprime integers $a,b$ ?
The representation $49=9^2-2\cdot 4^2$ shows that $d$ need not be squarefree.
these are numbers that are not divisible by $4$ or by any prime $q \equiv 3,5 \pmod 8.$ You also are throwing out the single factor of $2$ that would otherwise be allowed.