I figured out a interest property when trying to solve:
Create a sequence such that $K,a,b$ is an integer and the Greatest Common Denominator of $a \hspace{2mm}\&\hspace{2mm} b=1$; $$K=(\frac{a^2+b^2}{2})^{1/2}$$
I began solving this by creating algorithm that produces a list for all of the possible "$a$" values over a given range.
List of $a$ in range $(0,450)$ from my Python Algorithm
1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 113, 119, 127, 161, 167, 191, 199, 217, 223, 233, 241, 281, 287, 329, 337, 343, 353, 359, 383, 391, 433, 439, 449
I then started looking up similar sequence and found A253408 :
Find $x,y,z$ such that each variable is a integer. $$z^2+2=y^2+x^2$$
List of $(z-y)$ in range $(0,450)$ from A253408
1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113, 119, 127, 137, 151, 161, 167, 191, 193, 199, 217, 223, 233, 239, 241, 257, 263, 271, 281, 287, 289, 311, 313, 329, 337, 343, 353, 359, 367, 383, 391, 401, 409, 431, 433, 439, 449
Upon comparing these two I noticed that all of the numbers in the list of $a$'s exist in list for number in $(z-y)$'s or $a\subseteq (z-y)$. My first question is why does this work?
Also.
$z^2+2=x^2+y^2$, so $(z-y)=\frac{x^2-2}{(z+y)}$
$K=(\frac{a^2+b^2}{2})^{1/2}$, so $a=(2K^2-b^2)^{1/2}$
This means that if $a \subseteq(z-y)$ is True. Then $(2K^2-b^2)^{1/2}\subseteq\frac{x^2-2}{(z+y)}$ must also be true
My second question is, Is $(2K^2-b^2)^{1/2}\subseteq\frac{x^2-2}{(z+y)}$ a known equation?
Without doing all the work for you - where would be the fun in that! -
Looking at this modulo 4 casts a lot of light on the situation.
For any non-negative integers $x,y,z$:
$x^2\equiv 0$ or $1\pmod 4$
So $x^2+y^2\equiv 0,1$ or $2\pmod 4$,
and $z^2+2\equiv 2$ or $3\pmod 4$
So for $x^2+y^2$ to equal $z^2+2$, both sides must be $\equiv 2\pmod 4$.
So $x,y,x^2$ and $y^2$ must all be $\equiv 1\pmod 4$.
And $z^2$ must be $\equiv 0 \pmod 4$.
Also, any $K^2$ is $\equiv 0$ or $1\pmod 4$, so any $2K^2$ must be $\equiv 0$ or $2 \pmod 4$.
So if $a^2+b^2=2K^2$, $a,b,a^2$ and $b^2$ must all be either $\equiv0\pmod 4$ or all $\equiv 1 \pmod 4$. You've specified that they're odd, so they're all $\equiv 1\pmod 4$. So $K$ and $K^2$ are $\equiv 1\pmod 4$.
Etc. The two systems share many of the same properties.
p.s. It took me a while to realize what you were talking about; it's written in an atrociously bad style!