using the expansion in powers of $\Delta$ {#expansiont_0}
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By inserting the series (\[rozvojomeg0\]), (\[rozvojcalH0\]) into the first field equation (\[Eq1\]), the following key relation is obtained $$\begin{aligned}
&\sum_{l=2n-2}^{\infty}\Delta^{l}\sum^{l-2n+2}_{i=0}a_i\, a_{l-i-2n+2}\,(l-i-n+2)(l-3i-3n+1) \nonumber \\
& \hspace{35.0mm}=\tfrac{1}{3}k \sum^{\infty}_{l=p-4}\Delta^{l}\,c_{l-p+4}\,(l+4)(l+3)(l+2)(l+1) \,.
\label{KeyEq1}\end{aligned}$$ The second field equation (\[Eq2\]) puts further constraints on the admitted solutions, namely $$\begin{aligned}
&\sum_{l=2n+p-2}^{\infty}\Delta^{l}\sum^{l-2n-p+2}_{j=0}\sum^{j}_{i=0}a_i\,a_{j-i}\,c_{l-j-2n-p+2}\,(j-i+n)(l-j+3i+n+2)
+\sum_{l=2n}^{\infty}\Delta^{l}\sum^{l-2n}_{i=0}a_i\,a_{l-i-2n}
\nonumber \\
& = \tfrac{1}{3}k \bigg[2+\sum^{\infty}_{l=2p-4}\Delta^{l}\sum^{l-2p+4}_{i=0}c_{i}\,c_{l-i-2p+4}\,(i+p)(l-i-p+4)(l-i-p+3)(l-\tfrac{3}{2}i-\tfrac{3}{2}p+\tfrac{5}{2})\bigg]\,.
\label{KeyEq2}\end{aligned}$$ A considerably simpler is the additional (necessary but not sufficient) condition following from the trace equation (\[trace\]) which reads $$\begin{aligned}
&\sum_{l=n+p-2}^{\infty}\Delta^{l}\sum^{l-n-p+2}_{i=0}c_i\,a_{l-i-n-p+2}\,\big[(l-i-p+2)(l+1)+\tfrac{1}{6}(i+p)(i+p-1)\big] =-\tfrac{1}{3}\sum^{\infty}_{l=n}\Delta^{l}\,a_{l-n}
\,.
\label{KeyEq3}\end{aligned}$$

Now we analyze the consequences of the equations (\[KeyEq1\])–(\[KeyEq3\]).

First, by comparing the corresponding coefficients of the same powers of $\Delta^l$ on both sides of the key relation (\[KeyEq1\]), we can express the coefficients $c_j$ in terms of (products of) $a_j$. Moreover, the *terms with the lowest order* put further restrictions. In particular, comparing the lowest orders on both sides (that is ${l=2n-2}$ and ${l=p-4}$) it is obvious that *we have to discuss three distinct cases*, namely:

-   **Case I**: ${\ \ 2n-2<p-4}$,  i.e., ${\ p>2n+2}$,

-   **Case II**: ${\ 2n-2>p-4}$,  i.e., ${\ p<2n+2}$,

-   **Case III**: ${2n-2=p-4}$,  i.e., ${\ p=2n+2}$.

Now let us systematically derive all possible solutions in these three distinct cases.

**Case I**
