Q:

what is squared of a Kronecker ij?

Is it right to write $\delta_{ij}\delta_{ij}=(\delta_{ij})^2=\delta_{ij}$?

A:

Since $\delta_{ij}\in\{0,\,1\}$, it solves $x^2=x$. But you have to make explicit to the reader that $\delta_{ij}\delta_{ij}$ doesn't intend summation over repeated indices. (Even with $\delta_{ij}^2$, readers may think you intend such summation, especially if they're physicists.) If you sum over $i$, and each index has the same $n$ possible values, the result is $\sum_i\delta_{ij}^2=\delta_{jj}=1$; and if we also sum over $j$, we get $n$.

