How to find the number of right-angled triangles with integer sides and inradius $2009$? 
Problem: How to find the number of right-angled triangles with integer sides and inradius $2009$.

Please help, as I have no clue how to proceed with this problem. I do know that the inradius of a right-angle triangle with sides $a$, $b$, and $c$ is given by
$$r = \frac{ab}{a+b+c}.$$
 A: We also know that $a=k(u^2-v^2)$, $b=2kuv$, $c=k(u^2+v^2)$ for some integers $k,u,v$, so
$$ 2k^2(u^2-v^2)uv=2009k\cdot(u^2-v^2+2uv+u^2+v^2)$$
i.e. 
$$ k(u-v)v = 2009.$$
Now match the factors on the left with factors of $2009$.
A: I assume you are asking "how many primitive Pythagorean triples have inradius $r$ equal to 2009?" The answer depends on how many way we can write the factors $r=q'q$ where $q'$ is odd and where $q',q$ are coprime. Obviously, we can write $q'=1, q= 2009$ or as $q'=2009, q=1$. Two ways.
Note also that $r=2009$ has prime factors $7^2$ and $41$. Therefore, there are two more ways to write the inradius as the product of the 2 factors $r=q'q$ as defined above. Either $q'=41$ or $q'=7^2=49$.
The generalized Fibonacci sequence $[q',q,p,p']$ can be used to determine the integer sides of these four right triangles $(a,b,c)$ with inradius $r= q'q$.
One example is 
$(q',q,p,p')=(49,41,90,131)$. This corresponds to $(a,b,c)$ where
$r=q'q= (49*41)=2009$
$a=q'p'=6419$
$b=2qp=7380$
$c=pp'-qq'= 9781$.
Using the equations in the example above we can also determine the other 3 triples using the generalized Fibonacci sequences
$(q',q,p,p')=(41,49,90,139)$ yields $(a,b,c)=(5699, 8820,10501)$
$(q',q,p,p')=(1,2009,2010,4019)$ yields $(a,b,c)=(4019, 8076180, 8076181)$
$(q',q,p,p')=(2009,1,2010,2011)$ yields $(a,b,c)=(4040099, 4020,4040101)$
