If the inradius=$2013$ of a right angled triangle with integer sides. Find the no. of possible right angled triangles that can be formed using the above information. I have tried $r(a+b+c)=ab$ and $a^2+b^2=c^2$ , but couldn't reach further. Thanks in anticipation
One thing we can get is
$CF=CD=b-r$ and $BE=BD=c-r$. So, $BC=a=b+c-2r\quad (1)$.
By Euclide's formula we can look for a solution such that $\gcd(a,b,c)=1$ and $b=m^2-n^2$, $c=2mn$ and $a=m^2+n^2$ with, $\gcd(m,n)=1$ and not both odd.
We can then back to $(1)$ and get
$$a+2\cdot2013=b+c\\ m^2+n^2+2\cdot2013=m^2-n^2+2mn\\ n^2+2013=mn\to n(m-n)=2013$$
so, $n|2013$ and once $2013=3\cdot11\cdot61$ then we have the possibilities:
So you just have to try all possibilities for $n$ and find a suitable $m$.
For example, for $n=3$ you will find $m=674$ and then you have one solution which is $(a,b,c)=(674^2+3^2,674^2-3^2,2\cdot3\cdot674)$.
Now just find the others.
Answer to the question is easy: 8 primitive Pythagorean triples are there for the inradius 2013. There are no other triples (non-primitive Pythagorean triples).
My extended Answer to the question: 2013 has 3 prime factors such that 2013 = 3*11*61, thus, number of Pythagorean triples with inradius = 2^3 = 8. Therefore, there are 8 Pythagorean triples with the given inradius 2013. Ironically, all of them are primitive and no non-primitive Pythagorean triples present for the inradius of 2013.
Reference: Neville Robbins, On the number of primitive Pythagorean triangles with a given inradius, Fibonacci Quarterly 2006, 44(4), pp. 368–369.
For total Pythagorean triples, read: Tron Omland, How many Pythagorean triples with a given inradius?, Journal of Number Theory 2017, 170(1), 1–2.