A variation of Euclid's formula seems to reveal why primitive Pythagorean triple leg differences appear to be restricted to the number $\,1,\,$ to selected primes raised to any integer power, or to products of those primes, . Is my logic sound in the reasoning below?
\begin{align*}
&A=(2n-1+k)^2-k^2&&=(2n-1)^2+2(2n-1)k\\
&B=2(2n-1+k)k &&=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\
&C=(2n-1+k)^2+k^2 &&=(2n-1)^2+2(2n-1)k+2k^2
\end{align*}
$$B-A=\pm (j^2-2k^2),\,\, j=(2n-1), k,n\in\mathbb{N}$$
$$j^2\in\big\{1,9,25,49,81,\cdots\big\}\qquad
2k^2\in\big\{2,8,18,32,50,\cdots\big\}$$
\begin{align*}
(2n-1)^2\equiv\pm 1\pmod8\\
\,\land\quad 2k^2\equiv 0,4\pmod8\\
\implies (2n-1)^2-2k^2\equiv \pm1\pmod8
\end{align*}
$$\text{We are given that }\quad GCD(j,k)=1\quad
\text{which reduces the number of (j,k) commbinatons.}$$
$$\therefore \big\vert(B-A)\big\vert
\in\big\{1,7, 17, 23, 31, 41, 47, 49
,\cdots\big\}\equiv \pm 1\pmod 8$$
$\textbf{Update: }$ A comment from Keith Backman allowed me to insert the arguments just before the conclusion above. Is the logic complete now or does it need more proof?
Best Answer
I'm hoping to produce proofs of a couple of facts closely related to the topic of this question:
If $p$ is an odd prime dividing $u^2-2v^2$ with $\gcd(u,v)=1$, then $p\equiv\pm1\pmod8$.
If $p\equiv\pm1\pmod8$ is prime, then there exist integers $u,v$ such that $u^2-2v^2=p$.
This may take me a while, but the impatient reader should be able to find proofs in various introductory Number Theory textbooks, as well as on any number of websites.
Proof of 1. Let $p$ be an odd prime dividing $u^2-2v^2$ with $\gcd(u,v)=1$, so $u^2-2v^2\equiv0\pmod p$. If $p$ divides $v$, then it also divides $u$, contradiction, hence, $v$ is invertible modulo $p$, and we have $(uv^{-1})^2\equiv2\pmod p$, so $2$ is a quadratic residue modulo $p$. By the Second Supplement to the Law of Quadratic Reciprocity, this implies $p\equiv\pm1\pmod8$.
Now it may be objected that the Second Supplement to the Law of Quadratic Reciprocity, while quite well-known, and covered in every Number Theory textbook that does quadratic reciprocity, is not an obvious or immediate or intuitive result, so it should be proved here. I may get around to that, but for the time being I'll take it as read. Instead, let's make a brief diversion to show that $u^2-2v^2=15$ is impossible.
From $u^2-2v^2=15$ we derive $u^2-2v^2\equiv0\pmod3$. Now $2$ is not a quadratic residue modulo $3$, and you don't need the Second Supplement to see this, you just need to calculate $a^2\pmod3$ for $a=0,1,2$ to see that you never get $2$. So, the congruence has no solution with $v$ invertible modulo $3$. On the other hand, if $3$ divides $v$, then also $3$ divides $u^2$, so $3$ divides $u$, so $9$ divides $u^2-2v^2$, so $9$ divides $15$, contradiction. Hence, $u^2-2v^2=15$ is impossible.
For the proof of 2., I refer the reader to the answer by user Pablo at https://mathoverflow.net/questions/197918/what-is-known-about-primes-of-the-form-x2-2y2