Explain this n-modulus relation in Pythagorean triplets {a,a+n,c}

Question

I've found a seemingly consistent Pythagorean triplet relation for which I cannot find a proof or counterexample.

Theorem: For all Pythagorean triplets {a, b, c} where $gcd(a,b)=1$, b-a is odd and 2 is a square mod b-a.

To put that another way, for any Pythagorean triplet without common factors, $$b-a\in\{1,7,17,23,31,41,47,49,71,73,79,...\}$$

Example 1: 5 does not meet the critera $mod(k^{2},5)=2$, and we will find that for any triplet where b=a+5, a is a multiple of 5.

Example 2: $mod(10^{2},49)=2$ and we indeed find triplets do exist in which gcd(a,b)=1 and b=a+49, such as {11,11+49,61} and {35,35+49,91}

Is this theorem correct? Can anyone find a proof or counterexample?

Note: I'm not sure if this is useful or important, but I've also noticed that every number in the list is either prime or a product of primes in the list.

Answer 1

Your primitive Pythagorean triple is either $(2rs,r^2-s^2,r^2+s^2)$ or $(r^2-s^2,2rs,r^2+s^2)$ where $r+s$ is odd and $\gcd(r,s)=1$. Then $$b-a=\pm(r^2-2rs-s^2)=\pm[(r-s)^2-2s^2].$$ We see that $s$ is coprime to $b-a$, so $[(r-s)s']^2\equiv2\pmod {b-a}$ where $ss'\equiv1\pmod {b-a}$.

Answer 2

This formula generates triples where $\,GCD\,(A,B,C)=(2n-1)^2, \, n\in\mathbb{N}.\quad$ This includes all primitives where $\,GCD\,(A,B,C)=1^2.$

\begin{align*} &A=(2n-1)^2+2(2n-1)k\\ &B=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}

Subtracting, we get $\,C-B=\pm\,\big(j^2 - 2 k^2\big),\,\, j=(2n-1).\quad$ This means

$\,j^2\in\big\{9,25,49,81,121,\cdots\big\}\quad 2k^2\in\big\{2,8,18,32,50,\cdots\big\}.\quad$ Subracting one set's element from any of the other set's elements yields $\quad B-A\equiv \pm1\pmod8$

$\therefore\,$The difference for primitive triple legs is always $\,1,\,$ or a prime, or the product of primes of this form raised to any integer power –– not all primes are included. Of the 168 primes under 1000, only 80 are represented. Under 100, the differences are $1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97.$

Using these features we find that your conjecture must be true for some combination of $\,\big(A,A+(B-A),C\big),\,$ or $\big(B+(A-B),B,C\big)$

For $\,(A<B)$

$(3,3+1,5),\quad(5,5+7,13),\quad(13,13+71,85),\quad\cdots$

For $\,(A>B)$

$(8+7,8,17),\quad (20+1,20,29),\quad (12+23,12,37) ,\quad\cdots$