How to find all primitive triples (a,b,c)? (Pythagorean Triples) how to find all primitive triples when one value of $(a,b,c)$ is given? For example in this case $a = 45$.
What is the procedure to find the primitive triples ? 
Conditions for primitive triples are:


*

*$a^2 + b^2 = c^2$

*$(a,b) = 1$

*$a$ and $c$ are odd and $b$ is even.

We can find Pythagorean triples for a given side $A$ by solving Euclid's formula for $n$ and then testing a finite number of $m$-values to see which generates an integer for $n$.
$$A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad  \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil$$
The lower limit ensures $m^2>A$ and the upper limit ensures $m-n\ge 1$.
In this case, we have $n=\sqrt{m^2-45}\quad \text{where}\quad \lceil\sqrt{46}\space\rceil=7\le m \le \lceil 22.5\rceil=23$
Texting, we find $3$ values of $m$ in this range that yield integers for $n$. 
$\sqrt{7^2-45}=4,\quad F(7,4)=(33,56,65)\qquad GCD(33,56,65)=1$
$\sqrt{9^2-45}=6, \quad F(9,6)=(45,108,117)\qquad GCD(45,108,117)=9$
$\sqrt{23^2-45}=22, \quad F(23,22)=(45,1012,1013)\qquad GCD(45,1012,1013)=1$
We can see that $1$ of the triangles found not primitive. No other values of $m$ yield an integer so no other primitive triples exist where side $A=45$.
