A right triangle with integer sides has area equal to twice its perimeter. Find sum of all possible circumradii. 
In right triangle $ABC$, the area is twice the perimeter, and all sides have integer lengths. Compute the sum of all possible circumradii of $ABC$. 

I only have set up an equation 
$$\frac{xy}{2}=2\left(x+y+\sqrt{x^2+y^2}\right)$$ and 
$$R=\frac{xy \sqrt{x^2+y^2}}{2xy}=\frac{\sqrt{x^2+y^2}}{2}$$ 
 A: We can parameterize the sides of a right triangle ABC right-angled at C with integer sides in the following manner:
$$a=k(x^2-y^2),
b=2kxy,
c=k(x^2+y^2)$$
where k is any positive integer, x and y are co-prime integers with $x\not\equiv y\ (\textrm{mod}\ 2)$.
Using the condition, $\frac{ab}{2}=2(a+b+c)$, we obtain $ky(x-y)=4$ whence $k=4, y=1, x=2$ or $k=1,y=4,x=5$ or $k=2, y=2, x=3$. This leads to the triangles $(12,16,20)$, $(10,24,26)$ and $(9,40,41)$. In a right triangle, circumradius is half the hypotenuse. Therefore, $$ R=10,13, \text{or}\ 20.5 $$
A: $$\frac{xy}{2} = 2\left(x+y +\sqrt{x^2+y^2}\right) \implies (xy - 4(x+y))^2 = 16(x^2+y^2)\\ \iff xy(xy-8(x+y)+32) = 0 \implies (x-8)(y-8) = 32$$
Since $x$ and $y$ are integers and $32 = 2^5$, the total numbers of factors are $(5+1)=6$ which are $1,2,4,8,16,32$ .
So possible cases avoiding any redundancy for we just want circumradii are as follows:
$$\underset{32}{(x-8)}\underset{1}{(y-8)} = 32$$
$$\underset{16}{(x-8)}\underset{2}{(y-8)} = 32$$
$$\underset{8}{(x-8)}\underset{4}{(y-8)} = 32$$
So possible triplets turn out to be $40,9,41$;
$24,10,26$;
$16,12,20$.
So, possible circumradii are $\frac{41}2,\frac{26}2,\frac{20}2$.
