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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.

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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 $ss'\equiv1\pmod {b-a}$.

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