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Following the result of Ivan Niven, on 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 $\mathbb{Q}[\sqrt{-7}]$. Can someone show me the representation?

Following Jyrki Lahtonen's Answer, I have edited this question (adding condition $d=4k+3$).

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hmmmm... you cannot have two of the $s,t,u$ real as $4\cdot n^2\cdot 7-1\equiv 3\mod 4$ for all $n\in\mathbb Z$ hence cannot be written as a sum of $2$ squares. –  Alex Becker Feb 28 '12 at 12:37
    
Wait, note that since $7\equiv 3\mod 4$ the integers of $\mathbb Q[\sqrt{-7}]$ are of the form $a/2+b\sqrt{-7}/2$. –  Alex Becker Feb 28 '12 at 12:44

3 Answers 3

up vote 7 down vote accepted

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.

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Mr Lahtonen, I think Niven gives the condition for $d$, that is $d \equiv 3 mod 4$, which is satisfied for $d=7$. I have edited my question. –  Ajat Adriansyah Feb 28 '12 at 13:08
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@Ajat: My argument makes Niven's claim suspect for the rings of integers of all the fields $\mathbb{Q}[\sqrt{-d}]$ for $d=7,23,31,47,\ldots$. Have you had any luck with those? –  Jyrki Lahtonen Feb 28 '12 at 14:10
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Indeed $(1+\sqrt{-7})/2 \mapsto 3 \in \mathbb{Z}/8\mathbb{Z}$ extends to a homomorpshism on the ring of integers of $\mathbb{Q}(\sqrt{-7})$. –  WimC Feb 28 '12 at 15:48
    
@JyrkiLahtonen: I'd say all $d \equiv -1$ (mod 8) are suspect since $X^2-X+(d+1)/4$ has a root in $\mathbb{Z}/8\mathbb{Z}$ by Hensel lifting. –  WimC Feb 28 '12 at 17:00
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@WimC: For the benefit of those reader who are not familiar with the 2-adics you could consider writing the details of the use of a homomorphism to $\mathbf{Z}/8\mathbf{Z}$ as an answer! Most of us see it from the comment, but I'm sure many would prefer to see this settled that way! –  Jyrki Lahtonen Feb 28 '12 at 19:35

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."

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thank you very much for the reference :) –  Ajat Adriansyah Feb 29 '12 at 1:49
    
+1: Nice digging, Gerry. Wish I knew how to do that :-( –  Jyrki Lahtonen Feb 29 '12 at 11:05
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@Jyrki, I think I just went to the review of Niven's paper at Math Reviews and then clicked on the link to other papers/reviews that referenced Niven's paper. –  Gerry Myerson Feb 29 '12 at 21:43

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 $a^2+b^2+c^2 \neq -1$.

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+1 Sorry about being late with my upvote. I forgot about this in a two days or something. –  Jyrki Lahtonen Apr 30 '12 at 9:40

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