a dilogarithm identity: known or new? I was playing around with dilogarithms and numerically found the following dilogarithm identity:
$$\text{Li}_2\left(\frac{2 m}{m^2+m-\sqrt{((m-3) m+1)
   \left(m^2+m+1\right)}-1}\right)$$
$$-\text{Li}_2\left(\frac{m^2+m-\sqrt{((m-3) m+1)
   \left(m^2+m+1\right)}-1}{2 m^2}\right)$$
$$+\text{Li}_2\left(\frac{2 m^2}{(m-1)
   m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1}\right)$$
$$-\text{Li}_2\left(\frac{1}{2}
   \left((m-1) m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1\right)\right)$$
$$-\log(m)\log \left(-m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1\right)$$
$$+2\log(m)\log
   \left(\frac{m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1}{m^{3/2}}\right)$$
$$+\log(m)\left(\log(m)+i\pi -\log(2)\right)=0$$
where m is a real number in a neighborhood of 1 (such that the square root is real).
For those who use mathematica, please copy below
expr = Log[
    m] (I [Pi] - Log[2] + Log[m] - 
     Log[-1 + m - m^2 - Sqrt[(1 + (-3 + m) m) (1 + m + m^2)]] + 
     2 Log[(-1 + m + m^2 - Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])/m^(
       3/2)]) + 
  PolyLog[2, (
   2 m)/(-1 + m + m^2 - Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])] - 
  PolyLog[2, (-1 + m + m^2 - Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])/(
   2 m^2)] + 
  PolyLog[2, (2 m^2)/(
   1 + (-1 + m) m + Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])] - 
  PolyLog[2, 
   1/2 (1 + (-1 + m) m + Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])]
Does anybody have any idea how to prove this? 
 A: It looks to me like you don't even need the five-term identities.  If I denote the arguments of your dilogarithms by $a_1$, $a_2$, $a_3$ and $a_4$ I find that $(1-a_1) a_4 = 1$ and $a_2 a_3 = -1 + a3$.  Then I think using the simple formulas
$$
\text{Li}_2(z)=-\text{Li}_2(1-z)-\log (1-z) \log (z)+\frac{\pi ^2}{6},
$$ and
$$
   \text{Li}_2(z)=-\text{Li}_2\left(\frac{1}{z}\right)-\frac{1}{2} \log
    ^2(z)-\frac{\pi ^2}{6}
$$ should be enough (but some care is needed when placing the branch cuts).
A: All functional (i.e., depending on a parameter) relations are known to be a consequence of the 5-term relation of the dilogarithmic function; see [D. Zagier, The dilogarithm function, in: Frontiers in Number Theory, Physics and Geometry II, P. Cartier, B. Julia, P. Moussa, P. Vanhove (eds.), Springer-Verlag, Berlin-Heidelberg-New York (2006), 3-65]. In your case, I would suggest to compare first the derivatives w.r.t. $m$ and then use the fact that your identity is true for a particular value of $m$ (for example, $m=0$). 
A: The last comment of Sidious Lord, above, is correct: in fact, $a_4=1/(1-a_1)$ and $a_3=1/(1-a_2)$. Therefore, the application of Euler's reflection formula ($\mathrm{Li}_2(z) \rightarrow \mathrm{Li}_2(1-z)$), followed by an inversion ($\mathrm{Li}_2(z) \rightarrow \mathrm{Li}_2(1/z)$), as suggested by Sidious, actually reduces the expression to a few logarithms, which then simplify to zero.
