$-1$ as the sum of three squares of algebraic integers in $\mathcal{O}_{\mathbb{Q}(\sqrt{-7})}$ Following the result of Ivan Niven, in his paper
http://www.ams.org/journals/tran/1940-048-03/S0002-9947-1940-0003000-5/S0002-9947-1940-0003000-5.pdf 
That is whenever $d\equiv 3 \pmod 4$, every algebraic integer of the form $a + 2b\sqrt{-d}$ can be expressed as the sum of three squares. Which means that $-1$ can be expressed as $s^2 + t^2 + u^2$ in $\mathcal{O}_{\mathbb{Q}(\sqrt{-7})}$. Can someone show me the representation? 

Following Jyrki Lahtonen's answer, I have edited this question (adding the condition $d = 4k + 3$).
 A: According to the review by E G Straus,
Eljoseph, Nathan,
On the representation of a number as a sum of squares,
Riveon Lematematika 7, (1954). 38–43, MR0058627 (15,401d) 
"points out an error in the proofs of Theorems 4 and 10 of I. Niven's paper, Trans. Amer. Math. Soc. 48, 405-417 (1940) [MR0003000 (2,147b)]. 
"He points out that instead of proving that every number $a+2b\sqrt{-m}$ ($a,b,m$ rational integers, $m\ge2$, square free) is the sum of squares of three integers in ${\bf Q}(\sqrt{-m})$, all that Niven's argument for Theorem 4 proves is: 
"Either $a+2b\sqrt{−m}=\alpha^2+\beta^2+\gamma^2$ or $a+2b\sqrt{−m}=\alpha^2-\beta^2-\gamma^2$ where $\alpha,\beta,\gamma$ are integers in ${\bf Q}(\sqrt{-m})$. 
"The author gives a valid counterexample to Theorem 10 and an invalid counter-example to Theorem 4. However Theorem 4 was disproved by C. L. Siegel [Ann. of Math. (2) 46, 313--339 (1945); MR0012630 (7,49b)], who gives an infinite number of imaginary quadratic fields, including ${\bf Q}(\sqrt{-7})$, in which 7 is not the sum of three squares." 
A: 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 $a^2+b^2+c^2 \neq -1$.
A: I don't think that this can be done. $-7\equiv 1\pmod8$, so there is a square root of $-7$ in the 2-adic field $\mathbf{Q}_2$. Therefore the ring of integers of $\mathbf{Q}[\sqrt{-7}]$ can be viewed as a subring of the 2-adic integers $\mathbf{Z}_2$. But the equation
$$
s^2+t^2+u^2=-1
$$
has no solutions with $s,u,t\in\mathbf{Z}_2$, because reducing such an equation modulo 8 contradicts the fact that $s^2,u^2,t^2$ are all congruent to either $0,1$ or $4$ modulo $8\mathbf{Z}_2$.
I don't have the time to check Niven's argument to see, whether he might have made a mistake somewhere. A priori it is more likely that I missed something than he would have done so.
