This answer explains an earlier comment in more detail, as requested by Jyrki Lahtonen. I'll write $\mathbf{Z}_8$ for $\mathbb{Z}/8\mathbb{Z}$. First, as Jyrki remarked, the equation $a^2+b^2+c^2=-1$ has no solution in $\mathbf{Z}_8$. This follows by inspection from the fact that $x^2 \in \{0,1,4\}$ for all $x\in \mathbf{Z}_8$. Now let $d$ be a square free positive integer such that $d \equiv 7 \ (\bmod 8)$. Then $d \equiv 3\ (\bmod 4)$ and I'll just state the fact that in this case the ring of integers $\mathcal{O}_d$ in $\mathbb{Q}(\sqrt{-d})$ is a lattice:
$ \mathcal{O}_d = \mathbb{Z} + \mathbb{Z} \alpha, \textrm{ where } \alpha =\frac{1+\sqrt{-d}}{2}. $
Define $k := (d+1)/4 \in 2\mathbb{Z}$. Then the minimum polynomial of $\alpha$ is $X^2-X+k$. The next step is to show that this polynomial has two roots in $\mathbf{Z}_{2^m}$ for all $m \geq 1$. This follows by induction.
First note that $0,1$ are both roots in $\mathbf{Z}_2$. Now suppose that $x \in \mathbf{Z}_{2^m}$ is a root, so $x^2-x+k=y2^m$ for some integer $y$. Then
$ (x+y2^m)^2-(x+y2^m)+k = xy2^{m+1} + y^22^{2m} \equiv 0\ (\bmod 2^{m+1}) $
and in particular the minimum polynomial has a root x' \in \mathbf{Z}_{2^{m+1}} such that x' \equiv x\ (\bmod 2^m). In fact closer inspection of this procedure shows that there are exactly two roots in $\mathbf{Z}_{2^m}$.
Fix a root $x \in \mathbf{Z}_8$ and define a map $\varphi: \mathcal{O}_d \rightarrow \mathbf{Z}_8$ by
$ \varphi: a + b \frac{1+\sqrt{-d}}{2} \mapsto a + bx. $
It is easy to check explicitly that $\varphi$ is in fact a homomorphism. Multiplication works out because both $x \in \mathbf{Z}_8$ and $\alpha \in \mathcal{O}_d$ satisfy an identical quadratic relation. Now we can conclude that for any $a,b,c \in \mathcal{O}_d$
$ \varphi(a^2+b^2+c^2) = \varphi(a)^2+\varphi(b)^2+\varphi(c)^2 \neq -1 $
and therefore $a^2+b^2+c^2 \neq -1$.