I suspect if we have ever met!

We take logarithms on both sides and get $\begin{aligned} x^{x^{1/2}} &= (x^x)^{1/2} \\ x^{1/2} \ln x &= \frac{1}{2}\ln x^x \\ \sqrt{x} \ln x &= \frac{x}{2}\ln x \\ \sqrt{x} \ln x - \frac{x}{2}\ln x &= \ln x \left( \sqrt{x}-\frac{x}{2}\right) = 0\\ \end{aligned}$ So, either $\ln x = 0 \Longrightarrow x = e^0 = 1$ or $\begin{aligned} \sqrt{x}-\frac{x}{2} &= 0 \\ \sqrt{x} &= \frac{x}{2} \\ x &= \frac{x^2}{4} \\ 0 &= x^2 - 4x \\ x &= 0 \mbox{ or } x = 4\end{aligned}$ We check our values, and find that $x = 0$ is the only unacceptable solution (since it gives $0^0$ on both sides).

Therefore, $x = 1$ or $x = 4$ . The sum is $\fbox{5}$ .

Taking logs on both sides: $\ x^{1/2} log(x) = \frac{1}{2} x log(x) \implies\ log(x) . (x-2 \sqrt{x} ) = 0 \therefore\ x = 0,4$

Let's assume that $x\ge 0$ and rewrite this equation :

${ x }^{ \sqrt { x } }={ ({ x }^{ \frac { 1 }{ 2 } }) }^{ x }$

${ x }^{ \sqrt { x } }={ ({ x }^{ \frac { 1 }{ 2 } }) }^{ x }$

${ x }^{ \sqrt { x } }={ (\sqrt { x } ) }^{ x }$

Raise both sides to the power of 2:

${ x }^{ 2\sqrt { x } }=x^{ x }$

At the first glance it's obviously that $x1 = 1$ ; Then :

$2\sqrt { x } =x$

Again raise both sides to the power of 2:

$4x =x^2$

$x^2-4x=0$

$x(x-4)=0$

Where $x2 = 0, x3 = 4$ .

But $x2=0$ doesn't satisfy, because $0^0$ is undefined.

So, we get $x3 + x1 = 5$ .