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 dilogarithm identity: known or new?
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).


