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.
4 Answers
$\begingroup$There is a well-known parametrization of primitive triples as $(n^2-m^2, 2mn, n^2+m^2)$ for appropriate values of $n,m$. So for your problem you should proceed by solving for $n$ and $m$ coprime such that $n^2-m^2 = (n+m)(n-m) = 45$.
$\endgroup$ 6 $\begingroup$You seem to be asking: "How do I find all primitive triples with a leg equal to a given number, say 45?".
A simple method is to use the "Fibonacci Box" shown below where $q',q$ are positive coprime integers with $q'$ odd, and where $q'+q=p$ and $q+p=p'$.
$$ \left[ {\begin{array}{*{20}{c}} q & q' \\ p & p' \end{array}} \right]$$
Then:
Inradius $r=q'q$ (First row product).
Odd leg $a=q'p'$ (Right column product).
Even leg $b=2qp$ (Left column product times two).
Hypotenuse $c= pp'-qq'$ (Difference of bottom row product and top row product).
For a triple with leg equal to 45, we find the factor pairs $q',p'$ that meet the conditions. These are $q',p'=1,45$ and $q',p'=5,9$.
So only 2 primitive triples with odd leg equal to 45 exist.
$ \left[ {\begin{array}{*{20}{c}} 2 & 5 \\ 7 & 9 \end{array}} \right]$ yields triple $(45, 28, 53)$.
$ \left[ {\begin{array}{*{20}{c}} 22 & 1 \\ 23 & 45 \end{array}} \right]$ yields triple $(45, 1012, 1013)$
$\endgroup$ $\begingroup$Well, think about the simplest pythagorean triple. It is $(3,4,5)$. So if you are given that $a=45$, divide $45$ by $3$ since $45$ is $a$ in your example and $3$ is a in a $(3,4,5)$ triangle. $45/3=15$. Ok, so now that you have done that, take the other sides in a $(3,4,5)$ triangle and multiply them by $15$. $4\cdot 15=60$ and $5\cdot15=75$. So that means that in your example, $a=45$, $b=60$, and $c=75$. Hope this helps. I at least think this is what you are asking about but I'm not sure.
$\endgroup$ 1 $\begingroup$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 $A=45$.
$\endgroup$
