I'm trying to find triplets of integer $(x, y)$ pairs - $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ - that satisfy the following equations:
$$ {x_1}^2 + {y_1}^2 = {x_2}^2 + {y_2}^2 = {x_3}^2 + {y_3}^2 \\ x_1 + x_2 + x_3 = 5 \\ y_1 + y_2 + y_3 = 0 \\ (x_1, y_1) \neq (x_2, y_2) \\ (x_1, y_1) \neq (x_3, y_3) \\ (x_2, y_2) \neq (x_3, y_3) $$ Currently, for each integer $c$ that can be written as the sum of two squares, I create a list of all possible integer pairs $(a, b)$ for which $a^2 + b^2 = c$ and then check all possible triplets from each list for validity (actually I only need to check all possible combinations of 2 pairs, but the idea is the same). The only constraints I've found are $c$ must be divisible by 5 and $a$ and $b$ must have different parity. This is pretty inefficient, as only a very small subset of possible values of $c$ produce valid triplets. Checking the validity of any given triplet is relatively easy, so I'm hoping someone can help me find further constraints on the values of $c$. It is also entirely possible that there is some other, more efficient method of finding these triplets. Either way, help would be much appreciated.
We can use functions derived from the Pythagorean theorem where $A^2+B^2=C^2.\quad$ One generator is Euclid's formula, shown here as $ \quad A=m^2-k^2\quad B=2mk \quad C=m^2+k^2$
All $C$-values are of the form $4n+1$ but not all $4n+1$ are valid $C$-values. A short list is shown here. You will notice, for $C\in \{65,85,145,185,205,221,265,305,325,365\}$, there are $2$ entries because those have $2$ different primitive Pythagorean triples each. For $C=1105$, there would be $4$.
We can find the $(m,k)$ pairs for triples for any given $C$-value by solving the $C$-function for $k$ and testing a defined range of $m$-values to see which yield integers. If none are found, there is no primitive or $2\times$ or square-multiple of a primitive for that $C$-value. Here is how we find triples.
\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\\ \text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor \end{equation} The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\\ \quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$ $$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
The number of "primitive" triples will be $\quad 2^{n-1}$ where $n$ is the number of prime factors of $C$ meaning there will never be $3$ and only $3$ primitive triples for any $C$-value. For example, with $C=65=5\times13$, there are $2^1=2$ primitive triples. There $are$ values that also have non-primitive corresponding triples such as $C=325$. The first is non-primitive but those triples are
$$f(15,10)=(125,300,325)\quad f(17,6)=(253,204,325)\quad f(18,1)=(323,36,325)$$ So we have $\qquad125^2+300^2\space =\space 253^2+204\space =\space 323^2+36^2\space =\space 325^2$
The combination $125+253-323\ne5\space$ but $\space 125+253-323=55$ if that helps (kidding). Dividing both $x$ and $y$ by $11$ would be valid but $300-204-36\ne0$ so this triple of triple will not work in any case.
There are $67$ C-values where $\quad C=4n+1\space\text{ for }\space 81\le n\le 11925\quad$ that have $3$ triples each. You would have to find them programmatically as I did but perhaps one of them satisfies your requirement $\quad x_2+x_2+x_3=5\quad\land\quad y_2+y_2+y_3=0$