I asked a [question](https://math.stackexchange.com/q/1373123/19661) at M.SE a couple of years ago about polylogarithms [$\!^{[1]}$](http://en.wikipedia.org/wiki/Polylogarithm)[$\!^{[2]}$](http://mathworld.wolfram.com/Polylogarithm.html)[$\!^{[3]}$](http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog)[$\!^{[4]}$](http://dlmf.nist.gov/25.12) where I conjectured
$$720\,\operatorname{Li}\_4\!\left(\tfrac12\right)-2160\,\operatorname{Li}\_4\!\left(\tfrac13\right)+2160\,\operatorname{Li}\_4\!\left(\tfrac23\right)+270\,\operatorname{Li}\_4\!\left(\tfrac14\right)+540\,\operatorname{Li}\_4\!\left(\tfrac34\right)+135\,\operatorname{Li}\_4\!\left(\tfrac19\right)\\
=19\pi^4+30\left(\pi^2-\beta^2\right)\left(10\alpha^2-12\alpha\beta+3\beta^2\right)-30\,\alpha^2\left(19\alpha^2-24\alpha\beta+8\beta^2\right)\!,\\
\text{where $\alpha=\ln2,\,\,\beta=\ln3$.}$$
I put a bounty on it recently, but it still remains unanswered. So, I decided to re-post it here. Also, I recently discovered several similar conjectures for $\operatorname{Li}\_5(z)$ involving powers of the golden ratio $\phi=\frac{1+\sqrt5}2$:
$$4455\,\operatorname{Li}\_5\left(\phi^{-2}\right)-1215\,\operatorname{Li}\_5\left(\phi^{-4}\right)-360\,\operatorname{Li}\_5\left(\phi^{-6}\right)+15\,\operatorname{Li}\_5\left(\phi^{-12}\right)\\
=3015\,\zeta(5)-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi$$




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$$45000\,\operatorname{Li}\_5\left(\phi^{-2}\right)-16875\,\operatorname{Li}\_5\left(\phi^{-4}\right)-144\,\operatorname{Li}\_5\left(\phi^{-10}\right)+9\,\operatorname{Li}\_5\left(\phi^{-20}\right)\\
=28944\,\zeta(5)-6000\ln^5\phi+1600\,\pi^2\ln^3\phi-356\,\pi^4\ln\phi$$




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$$15660\,\operatorname{Li}\_5\left(\phi^{-2}\right)-19440\,\operatorname{Li}\_5\left(\phi^{-4}\right)+7680\,\operatorname{Li}\_5\left(\phi^{-6}\right)+2430\,\operatorname{Li}\_5\left(\phi^{-8}\right)-15\,\operatorname{Li}\_5\left(\phi^{-24}\right)=5025\,\zeta(5)+2088\ln^5\phi-240\,\pi^2\ln^3\phi-28\,\pi^4\ln\phi$$


How can we prove these conjectures? Are there other similar identities of this kind?



Several conjectured identities for polylogarithms
***(Updated answer)***: 


Upon further research, it turns out your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "*[The Polylogarithm in Algebraic Number Fields](https://ac.els-cdn.com/0022314X85900526/1-s2.0-0022314X85900526-main.pdf?_tid=3b94c1e0-e19f-11e7-bdfc-00000aab0f6c&acdnat=1513346147_cef17635ac98773bf7c6d945f73cbde9)*" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.


**I. Ladder of index 12**


$$A = \frac{\text{Li}\_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}\_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}\_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}\_n(\rho^2)}{2^{n-1}}$$


$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$


then for $n=1,2,3,4,5$,


$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$


with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation. Incidentally, a consequence (*or is it the other way round?*) of this ladder is the high-degree cyclotomic equation satisfied by the golden ratio,


$$(1 - \phi^{-12}) = (1 - \phi^{-6})^{3/2}(1 - \phi^{-4})(1 - \phi^{-2})^{-11/48}\,\phi^{13/48}$$


Notice that the powers of $(1 - \phi^{k})$ are coefficients in $A,B$.


**II. Ladder of index 20**


Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.


**III. Ladder of index 24**


Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.




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***(Old answer)***: As to your question if there are similar identities, it seems there is a whole family,


$$5\,\text{Li}\_2(\phi^{-1}) 
=-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$




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$$15\,\text{Li}\_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$




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$$144\text{Li}\_4(\phi^{-1})-9\text{Li}\_4(\phi^{-2})-64\text{Li}\_4(\phi^{-3})+4\text{Li}\_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$




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$$4455\,\text{Li}\_5\left(\phi^{-2}\right)-1215\,\text{Li}\_5\left(\phi^{-4}\right)-360\,\text{Li}\_5\left(\phi^{-6}\right)+15\,\text{Li}\_5\left(\phi^{-12}\right)\\
=-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$




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$$m\_1\text{Li}\_6(\phi^{-2})+m\_2\text{Li}\_6(\phi^{-4})+m\_3\text{Li}\_6(\phi^{-6})+m\_4\text{Li}\_6(\phi^{-8})+m\_5\text{Li}\_6(\phi^{-12})+m\_6\text{Li}\_6(\phi^{-24})\\
=m\_7\ln^6\phi+m\_8\pi^2\ln^4\phi+m\_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$




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$$n\_1\text{Li}\_7(\phi^{-2})+n\_2\text{Li}\_7(\phi^{-4})+n\_3\text{Li}\_7(\phi^{-6})+n\_4\text{Li}\_7(\phi^{-8})+n\_5\text{Li}\_7(\phi^{-12})+n\_6\text{Li}\_7(\phi^{-24})\\
=n\_7\ln^7\phi+n\_8\pi^2\ln^5\phi+n\_9\pi^4\ln^3\phi+n\_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$




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and where the integers $m\_i$ and $n\_i$ were found using *Integer Relations* but are too tedious to write here.


*Does this go on forever*? Note the **exponents** of $\phi$ in $\text{Li}\_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$. 


**P.S.** Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.


