How to prove $tr\{\sigma_1\sigma_2\sigma_3\}=\pm 2i$ with only using $\{\sigma_i,\sigma_j\}=2\delta_{ij}$ How to prove $tr\{\sigma_1\sigma_2\sigma_3\}=\pm 2i$ with only using $\{\sigma_i,\sigma_j\}=2\delta_{ij}$ and we cannot use the explict representation of Pauli matrices and cannot use $[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k$.
There is quantum mechanics homework to ask us prove $tr\{\sigma_i\sigma_j\sigma_k\}=2i \epsilon_{ijk}$  and all things we can only use is only $\{\sigma_i,\sigma_j\}=2\delta_{ij}$. I can prove the indeces $ijk$ is antisymmetric but cannot solve $tr\{\sigma_1\sigma_2\sigma_3\}=2i$ with only $\{\sigma_i,\sigma_j\}=2\delta_{ij}$. 
Firstly, thanks to Valter Moretti's answer, this consequece is not correct in general. Because minus pauli matrices still satisfy the anticommutator relation but the trace of three is $-2i$
But I still have a hunch that there are only two possibilities of $2\times 2$ representation of an associative algebra generated by $\sigma_1, \sigma_2, \sigma_3$ with $\{\sigma_i,\sigma_j\}=2\delta_{ij}$ $i,j\in\{1,2,3\}$. One is similar to pauli matrices and the other one is similar to the minus pauli matrices. Can this claim be proven?
 A: It is impossible to prove the wanted identity (initial  version) making only use of the anticommutation relation. It is  because if you change the sign of all $\sigma_k$, the anticommutation relation remains satisfied but the trace of the product of three matrices changes its sign. The only thing you can at most prove, also exploiting the cyclic property of the trace, is that the afore-mentioned trace is proportional to $\epsilon_{ijk}$.
ADDENDUM
Regarding your last conjecture, I think it is correct provided the matrices are traceless and Hermitian as the matrices in the real  span if Pauli matrices are (so you should consider the real span of Pauli matrices instead of the associative complex algebra generated by them which is the whole algebra of 2×2 complex matrices). Indeed, consider the three traceless Hermitian   matrices $t_i$. Thus you can write 
$$t_i = R_{ij} \sigma_j$$
for real coefficients $R_{ij}$. Imposing anticommutation relations for the matrices $t_i$ and using the fact that Pauli matrices satisfy the same relations, you have
$$R_{ij}R_{kj} = \delta_{ik}.$$
So the matrices $R$ are in $O(3)$. If the determinant of $R$ is $1$, as is well known, there is an operator $U\in SU(2)$ such that, as you conjectured
$$t_i = U \sigma_i U^\dagger$$
for $i=1,2,3$.
If instead the determinant of $R$ is $-1$, then $-R$ has determinant $1$ and therefore there is a unitary matrix $V$ such that
$$t_i = -V \sigma_i V^\dagger$$
for $i=1,2,3$.
Using the cyclic property of the trace, this result also  proves your initial identity in the $\pm$ version.
ADDENDUM2.  
The traceless condition can be dropped. Consider three 2×2 Hermitian matrices $t_i$ satisfying Pauli matrices' anticommutation realtions. So it must be $t_kt_k=I$ for $k=1,2,3$ (no summation over repeated indices is supposed above). Therefore each $t_i$ is simultaneously Hermitian and unitary and thus its eigenvalues are in the set $\{\pm 1\}$. If both eigenvalus are taken the trace of the matrix is $0$, otherwise $t_i=\pm I$. This second case is incompatible with t he anticommutation relations:
$$ 0 = t_i t_k + t_k t_i = 2 t_k$$
which is in contradiction with $t_kt_k = I$.
We can therefore conclude that your conjecture is true assuming that the matrices are Hermitian: 
Theorem
If three 2×2 Hermitian matrices satisfy the anticommutation relations of Pauli matrices, then they can be simultaneously transformed into either the three Pauli matrices or the three Pauli matrices with negative sign by means of a common  change of orthonormal basis. The change of basis is performed by means of a matrix in $SU(2)$.
Very nice...
