How many right angled triangles with co-prime integer sides and base of length $28cm$ are there? Please help me.
My working: I tried assuming that once that base side is $2m = 28$ as $ 2m,m^2-1,m^2+1 $ form Pythagorean triplets. I got one set of possible answer which was $28, 195$ and $197$. Please help me to find the other sets.
We begin with Euclid's formula shown here as $$ \quad A=m^2-k^2 \qquad B=2mk \qquad C=m^2+k^2\quad$$ We can find all triples with $B=28$ by solving the $B$-function for $k$ and testing a defined range of $m$-values to see which, if any, yield integers.
\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 $m\ge 2$ $$B=28\implies\bigg\lfloor \frac{1+\sqrt{56+1}}{2}\bigg\rfloor =4 \le m \le \frac{28}{2}=14\\ \quad \text{and we find} \quad m\in\{7,14\}\implies k\in\{2,1\}$$ $$F(7,2)=(45,28,53)\qquad \qquad F(14,1)=(195,28,197)\qquad $$
If we go with non-primitives for side-$B$, we find $$7\times F(2,1)=7\times(3,4,5)=(21,28,35)$$ All $B$-values are multiples of $4$ so there are no other triples.
A similar technique for side-$A$ shows just one non-primitive triple $4\times F(4,3)=4\times(7,24,25)\space = \space F(8,6)=(28,96,100)$