Prove diophantine equation $S^2+R^2+(r_1-r_2)^2 = 2R(r_1+r_2)$ has at most one solution

Question

Given this diophantine equation:

$$S^2+R^2+(r_1-r_2)^2 = 2R(r_1+r_2)$$

$S,r_1,r_2$ are variables. $R$ is a given constant. all values are positive integers.

How do I prove that there's at most one solution, not counting solutions where $r_1$ and $r_2$ are exchanged.

Thanks.

Answer 1

But I found two solutions:

(1) $S=4R, r_1=5R, r_2=4R$

(2) $S=4R, r_1=5R, r_2=8R$

Answer 2

For the equation.

$$S^2+R^2+(x-y)^2=2R(x+y)$$

You can set some numbers infinitely different way.

$$R=(a-b)^2+(c-d)^2$$

Then decisions can be recorded.

$$S=2(cb-ad)$$

$$x=b^2+d^2$$

$$y=a^2+c^2$$