I studied which differences $a-b$ are possible in primitive pythagorean triples $(a/b/c)$.
I noticed that the difference $d$ must be odd and contains only prime factors with quadratic residue $2$, in other words primes of the form $8k\pm 1$.
My conjecture is that this condition is also sufficient. To prove that I need the following :
Every prime $p$ of the form $8k\pm 1$ can be written as $p=a^2-2b^2$ (Infinite descent might be a possibility to prove it)
If two numbers $m,n$ are of the form $a^2-2b^2$ with coprime integers $a,b$, then so is the product $mn$.
I found out the idendity $$(a^2-2b^2)(c^2-2d^2)=(ac+2bd)^2-2(ad+bc)^2$$ which could be helpful.