If x,y and z are positive integers and $\frac 1x + \frac 1y = \frac 1z$ then $\sqrt{x^2+y^2+z^2}$ is rational.

Question

To solve this problem I first started off by factoring to get $z^2=(x-z)(y-z)$ only to realise that this does nothing so I then tried squaring both sides to get the reciprocals of $x,y$ and $z$ squared minus $2$ over $xy$ but after this I have no idea what to do any form of help would be appreciated.(sorry for not using mathjax as my browser does not support it)

Answer 1

The solutions to $\dfrac 1x + \dfrac 1y = \dfrac 1z$ are characterized by (See this)

\begin{align} x &= kp(p+q) \\ y &= kq(p+q) \\ z &= kpq \end{align}

Then \begin{align} \dfrac{x^2+y^2+z^2}{k^2} &= p^2(p+q)^2 + q^2(p+q)^2 + p^2q^2 \\ &= p^4 + 2p^3q + 3p^2q^2 + 2pq^3 + q^4 \\ &= \left( \begin{array}{l} p^4 &+ &p^3q &+ &p^2q^2 &+ \\ & &p^3q &+ &p^2q^2 &+ &pq^3 &+ \\ & & &+ &p^2q^2 &+ &pq^3 &+ &q^4 \end{array} \right) \\ &= p^2(p^2 + pq + q^2) + pq(p^2 + pq + q^2) + q^2(p^2 + pq + q^2) \\ &= (p^2 + pq + q^2)^2 \end{align}

Answer 2

Because $x\neq-y$, $z=\frac{xy}{x+y}$, $x^2+xy+y^2>0$ and $$\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+\frac{x^2y^2}{(x+y)^2}}=$$ $$=\frac{\sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=\frac{x^2+xy+y^2}{x+y}.$$

Answer 3

$(x+y-z)^2=x^2+y^2+z^2+2xy-2xz-2yz=(x^2+y^2+z^2)+2xyz\underbrace{\left(\dfrac 1z-\dfrac 1y-\dfrac 1x\right)}_0$

So $\sqrt{x^2+y^2+z^2}=|x+y-z|$ so not only it is rational, but also integer.