$\frac{1}{a} = \frac{1}{b} + \frac{1}{c} - \frac{1}{abc}$ and $a^2 + b^2 = c^2$ I have found this in an Romanian magazine. We have to solve for natural numbers:
$$\frac{1}{a} = \frac{1}{b} + \frac{1}{c} - \frac{1}{abc}$$
$$a ^ 2 + b ^ 2 = c ^ 2$$
After some elementary calculations, we get that $bc + 1 = a(b + c)$. I tried using some divisibility things, hoping to get $b + c$ dividing some constant, but, in the end, got $b + c \ | \ (b + c) ^ 2$. Then, $a^2(b + c) ^ 2 = (bc + 1) ^ 2$, so $(c^2 - b^2)(b + c)^2  =(bc + 1) ^ 2$  and tried solving this, but, again, couldn't. Can you help me? I doubt there are natural solutions, but are there any integers?

(This is not a solution. I made an error.)
Show that  (If you get stuck, show your work.)

*

*$(a, b, c)$ is a primitive pythagorean triplet.

*If $(a, b, c ) = (m^2 - n^2, 2mn, m^2+n^2)$ with $m > n > 0$, then there are no solutions.


 Substituting in, we get $n^4 = m^4 - 4mn^3$.
 View this as a quartic in $ x = \frac{n}{m}$, we get $ x^4 + 4x^3$ - 1 = 0, which has no rational solutions.
 Hence, there are no integer solutions to $(m, n)$.



*If $(a, b, c) = ( 2mn, m^2 - n^2, m^2 + n^2)$ with  $m > n > 0$, then the only solution is $(m, n) = (4, 1) $ which leads to $ (a, b, c) = (8, 15, 17)$.

(Hm, I thought I had a solution to this part, but it turns out that I made an error.)

 Substituting in, we have $ m^4 - 4m^3 n= n^4 - 1$.
 If $ n = 1$, we have $ m = 4n = 4$.
 Otherwise, writing it as $ m^3 (m-4n) = n^4 - 1 > 0 \Rightarrow m -1 \geq 4n$.

