find all positive solutions of $\frac{\ln^2(x^2)}{x^2} = \pi^2$ This is a riddle I have to solve.
What is surprising is the "all positive solutions" .
Obviously this equation involves the Lambert function, but "positive' is meaningless for complex numbers...
I have currently 4 solutions, 0.4745409994, -0.4745409994, i and -i, but one single positive (real) solution.
However, the riddle solution must contain more that one value.
The problem is stated exactly like in the title of my question, not $\frac{\ln(x^2)}{x^2}^2 = \pi^2$
Could this be the trick ? $\ln(\ln(x)) = \ln^2(x)$ ?
 A: Without Lambert function.
Consider that you look for the zero's of  function
$$f(x)=\log ^2\left(x^2\right)-\pi ^2 x^2$$ Let $x^2=t$ to make it
$$g(t)=\log ^2\left(t\right)-\pi ^2 t\implies g'(t)=\frac{2 \log (t)}{t}-\pi ^2$$ Since $\forall t \,\,\, \log(t)<t$, then $ g'(t)< 0  \,\,\,\forall t$. If the is a root, it is unique and since $g(t)$ veries from $+\infty$ to $-\infty$, the root does exist.
A: The positive real solution you got seems to be unique. 
We can evaluate:
$$\ln^2(x^2)=(x\pi)^2\to \ln(x^2)=x\pi\to x^2=e^{\pi x}\to e^{x\pi}-x^2=0$$
Yes, I ignored negative roots here, but it doesn't matter here because the real roots will have the same absolute value.
What matters here is that $f(x)=e^{x\pi}-x^2\implies f'(x)=\pi e^{\pi x}-2x>0\forall x\in\Bbb R^+$ (and you can check the same for $e^{-\pi x}-x^2$ and $e^{\pi x}-x^{-2}$)  so the function is monotone and will only have one solution.
A: We can use the Lambert W function to solve for $x$.
$$\begin{align*} \left(\dfrac{\ln (x)}{x}\right)^2=\pi^2\end{align*}$$
\begin{array}{lll}%
&\ln(x) = x\pi  \quad &  \ln(x) = -x\pi  \\
\implies& x=\exp(-W(-\pi)) \qquad  \implies &= \exp(-W(\pi))
\end{array}
Hence the two solutions are $\exp(-W(-\pi))\approx-0.159-0.585i$ and $\exp(-W(\pi))\approx 0.342$
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Assuming that there is a typo, 
this  is a minimal correction,
with which the phrase  "all positive solutions"
is justified: 
\begin{align}
\tfrac1{x^2}\,\ln(x^2)^2 &= \tfrac1{\pi^2}
,\\
\tfrac1x\,\ln(x^2) &= \pm\tfrac1{\pi}
,\\ 
\tfrac1x\,\ln(x) &=\pm\tfrac1{2\,\pi}
,\\
-\tfrac1x\,\ln(x) &= \pm\tfrac1{2\,\pi}
,\\
\tfrac1x\,\ln(\tfrac1x) &= \pm\tfrac1{2\,\pi}
,\\
\ln(\tfrac1x)&=\W(\pm\tfrac1{2\,\pi})
,\\
\tfrac1x&=\exp( \W(\pm\tfrac1{2\,\pi}))
,\\
x&=\exp(-\W(\pm\tfrac1{2\,\pi}))
.
\end{align} 
And there are three distinct real solutions:
\begin{align} 
x&=\exp(-\W(\tfrac1{2\,\pi})) \approx 0.8706
,\\
x&=\exp(-\Wp(-\tfrac1{2\,\pi})) \approx 1.2129
,\\
x&=\exp(-\Wm(-\tfrac1{2\,\pi})) \approx 18.2460
.
\end{align} 
$\endgroup$
