Natural number matrix solutions to $\sigma_i\sigma_j+\sigma_j\sigma_i = I\delta_{ij}$ Given the two matrices:
$\sigma_i$ and $\sigma_j$
we can construct a Clifford algebra based on the anti commutator rule:
$$\{\sigma_i,\sigma_j\}=\delta_{ij}1$$
where $\delta_{ij}$ is the Kronecker symbol. The question is: if the matrices are $(N\times N)$ and their elements are Natural numbers, how many matrices vs. $N$ can I find satisfying the anti commutator equation? 
I would appreciate any suggestion.
 A: To briefly synthesize the comments by Willie Wong and Eric O. Korman above, there are no (natural number) solutions to the original equation you posted unless you meant $\{\sigma_i,\sigma_j\}=2I\delta_{ij}$. Eric's comment applies if we are talking about integer matrices, and it's true that once you find an integer matrix solution $(a,b)$, then $(xax^{-1},xbx^{-1})$ is a solution for every invertible integer matrix $x$, of which there are many. Some of the $x$'s are even filled with natural numbers, and as I found out at this MO post, they are all permutations.
My contribution is that even if you adopt the change to $2I\delta_{ij}$, you will still not have any natural number matrix solutions. 
If $\sigma_i$ and $\sigma_j$ both have natural number entries, then $\sigma_i\sigma_j$ and $\sigma_j\sigma_i$ also have natural number entries. But {$\sigma_i,\sigma_j\}=0$ implies $\sigma_i\sigma_j=-\sigma_j\sigma_i$. The left hand side has nonnegative entries and the right hand side has nonpositive entries, so both sides are the zero matrices. 
But then $\sigma_i\sigma_j=[0]$ implies $[0]=\sigma_i\sigma_i\sigma_j\sigma_j=2I\cdot 2I=4I$, a contradiction. So, there are zero natural number matrix solutions.
