**Statement of the problem** 

I need to solve the equation 

\begin{align} 
0 = \frac{1}{\phi} \partial_{\sigma}\partial_{\sigma} \phi \hspace{20mm} (1)
\end{align}

where $\phi$ is a scalar field and we're working in Euclidean 4-space. More specific questions are asked below. Basic advice on this would also be appreciated.

**Context**

I'm trying to work through 't Hooft's Euclidean Yang-Mills N-instanton solution. The idea is that we make the ansatz for the gauge potentials

\begin{align}
A_{\mu} =  i \overline{\sigma}_{\mu \nu} \partial^{\nu}(\ln \phi(x))
\end{align}

where $\overline{\sigma}_{\mu \nu}$ can be thought of as a constant 4x4 matrix that is unimportant to my question. We use the self-duality of the field strength tensor that arises from this ansatz to obtain the equation (1) for $\phi$

**The Difficulties**

One generally good source has been Rajaraman's "Solitons and Instantons." The relevant section is displayed

![Relevant section in Rajaraman][1]


  [1]: https://i.sstatic.net/mdLWX.png

My difficulties are:

1. I am assuming by "singular" Rajaraman means that the function contains singularities. Why, explicitly, does $\phi$ being non-singular imply that we we need only solve $\Box \phi = 0$? Surely if $\phi$ contains any zeros then $1/\phi$ raises issues?
2. Why does $\Box \phi = 0$ only permit the constant solution and not say $\phi(x) = a_{\mu} x^{\mu} + b$ where $a_{\mu}$ and $b$ are constants? Is this because of the form of the ansatz, or is it something I'm missing?
3. In the case where $\phi$ is allowed to be singular, why does $\phi = 1/|x|^2$ solve the equation (1) for $x \neq 0$? Surely we have

\begin{align}
\frac{1}{\phi} \partial_{\sigma}\partial_{\sigma} \phi 
&= |x|^2 \partial_{\sigma}\partial_{\sigma}  \frac{1}{|x|^2}\\
&= |x|^2 \partial_{\sigma}\bigg(-\frac{2 x_{\sigma}}{|x|^4}\bigg)\\
&=|x|^2\partial_{\sigma} (\frac{-2 x_{\sigma}}{|x|^4})\\
&= |x|^2\partial_{\sigma} (\frac{-2 x_{\sigma}}{(\sum_{i} x_i^2)^2})\\
&= |x|^2\frac{-2}{(\sum_i x_i^2)^2} + |x|^2\frac{8 x_{\sigma}^2}{(\sum_i x_i^2)^3}\\
&=\frac{-2}{|x|^2} + \frac{8 x_{\sigma}^2}{|x|^4}\\
&\neq 0
\end{align}