Why Pythagorean triple leg differences have only certain values.

Question

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?

Answer 1

I'm hoping to produce proofs of a couple of facts closely related to the topic of this question:

1. If $p$ is an odd prime dividing $u^2-2v^2$ with $\gcd(u,v)=1$, then $p\equiv\pm1\pmod8$.

2. 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