Can we guess the condition 
$\Bbb Z_p[i]$ is a field is in general not true. 

I tried for $p=2,5,13$ and for these it is not a field, whereas for $p=3,7,11$ it is a field. What can we say about $p$ ? I think $p\equiv 3\pmod 4$.
 A: I'm not sure exactly what you read, but let's prove what I claim. I'll use $\Bbb Z_p$ to denote the integers modulo $p$ and $R[x]$ to denote the ring of polynomials in $x$ over $R$ and $i$ to be something satisfying $i^2=-1$.
Then in the case $p\equiv 1\mod 4$ we have that there is an integer $n$ which, when reduced and viewed as an element of $\Bbb Z_p$, squares to the equivalence class of $-1$ modulo $p$. This follows from Gauss' two squares theorem.
Now, for $p=2$ clearly $1^2=1\equiv -1\mod 2$, so $i=1$ in this case.
Finally for $p\equiv 3\mod 4$, we have that there is nothing in $\Bbb Z_p$ which squares to $-1$. As a result $x^2+1$ has no roots modulo $p$ and is so an irreducible polynomial in the ring $\Bbb Z_p[x]$. Since this is a Euclidean domain (it's a polynomial ring over a field) irreducibles generate maximal ideals, and so
$$\Bbb Z[x]/(x^2+1)$$
is a field. Furthermore, since the equivalence class of $x$ squares to the equivalence class of $-1$ in this quotient, $[x]=i$ in this ring (or $[-x]$ if you prefer).
So this ring which is $\Bbb Z_p[i]$ is isomorphic to a field, hence is a field.
