How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.

Using Euclid's formula, we can find triples matching any hypotenuse by solving the C-function for k and seeing which of a range of m-values yield integers. The limitation is that, for primitive triples, C must take the form $C-4n+1$ and not all of these are valid. Some valid ones are $5,13,17,25,29,37,41 ... 65$
\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 $k\in\mathbb{N}$.
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
There is a side-A matching every odd number greater that one.
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad  \sqrt{A+1} \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=15\implies \sqrt{15+1}=4\le m \le \frac{15+1}{2} =8\quad\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$
There is a side-B for every multiple of four.
\begin{equation}
B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$
$$B=44\implies\qquad \bigg\lfloor \frac{1+\sqrt{88+1}}{2}\bigg\rfloor =5 \le m \le \frac{44}{2}=22\quad \land \quad m\in\{11,22\}\implies k\in\{2,1\}$$
$$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)$$
