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In a given non-real algebraic number field (say, given by an irreducible polynomial over $\mathbb{Q}$) is there a complexity bound on the summands $x,\dots,t$ that make $-1$ a sum of squares? So $x^2+\dots+t^2=-1$.

Can you show that if $-1$ is a sum of squares at all, then it is a sum of some list of summands subject to some bound?

I would like to learn of an exponential bound. Say, an exponential function of the Weil heights of the rational coefficients of the defining polynomial, which serves as upper bound on the Weil heights of the rational coefficients of $x,\dots,t$ in the power base by the variable of the defining polynomial.

Of course if some other representation is more useful for the problem that's great with me.

To be clear, I am not asking how many summands it takes (I chose $x\dots t$ to suggest the four $x,y,z,t$). I am asking how complex they need to be. The shortest sum might not have the simplest entries.

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Even though you are not asking "how many summands", it may be useful to note that the minimal number of summands is called the "stufe" of the field, and, if it exists, is 1, 2, or 4. This is a theorem of Siegel, and is given on page 381 of Lam, Introduction to Quadratic Forms over Fields, Graduate Studies in Mathematics 67, American Mathematical Society, according to – Gerry Myerson Mar 16 '14 at 22:37
But I believe stufe have not got to do with (necessarily) integral summands, and so do not directly address the question of complexity. – Colin McLarty Mar 17 '14 at 9:22
Nowhere in your question do I see any mention of integrality. – Gerry Myerson Mar 17 '14 at 11:35
@GerryMyerson The question concerns complexity. For integers, including algebraic integers, complexity agrees with magnitude, and so Raghavan's magnitude bound also bounds complexity. For rationals and other non-integer algebraic numbers very small can be very complex. – Colin McLarty Mar 17 '14 at 13:46
So, $99-70\sqrt2$, having a much smaller magnitude than $99+70\sqrt2$, also has a much smaller complexity? – Gerry Myerson Mar 17 '14 at 22:28
up vote 6 down vote accepted

To answer the comment above; sure. There's a theorem of Cassels, generalized to number fields by Raghavan "Bounds for minimal solutions of Diophantine equations," (MR0485681) which gives a concrete upper bound for the smallest solution to a quadratic form (when they exist).

Namely, given an anisotropic quadratic form, there's a solution in the ring of integers with a bound that depends only on the discriminant of the number field, the size of the coefficients, and the number of variables. In this case, because the quadratic form is fixed, the bound depends only on the discriminant, explicitly;

$$|x_i| \le 625 |\Delta_F|^{5/2[F:\mathbb{Q}]}$$

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Indeed this answers the question asked. – Colin McLarty Mar 16 '14 at 22:17

If $F$ is a number field of degree $d$, then it has $d$ embeddings into $\mathbb{C}$. The signature of $F$ is defined to be $(r,s)$, where $r$ of these embeddings land in $\mathbb{R}$, and $s$ pairs of complex conjugate embeddings lie in $\mathbb{C}$. A number theorist would say that $F$ is totally complex if $r = 0$. I will assume that this is what you mean by non-real, because $-1$ is certainly not a sum of squares in any field with at least one real embedding.

Assume that $F$ is totally complex. I claim that

$$x^2_1 + x^2_2 + x^2_3 + x^2_4 + x^2_5 = 0$$

will always have a (non-zero) solution in $F$. It will have solutions over all local completions $F_v$, because all quadratic forms in $\ge 5$ variables have solutions over all local fields. (Alternatively, it is easy to check directly that it has a solution in $\mathbb{Q}_p$ for all primes $p$ by Hensel's Lemma). On the other hand, by assumption, it also has a solution over all completions $F_v$ of $F$ at the infinite primes, because $F$ is totally complex and so $F_v = \mathbb{C}$. Hence, by the Hasse-Minkowski theorem, it has a solution over $F$. In particular, in such fields $F$, one can always write

$$ - 1 = a^2 + b^2 + c^2 + d^2$$

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Can you put some complexity bound on the summands $x_1,x_2,x_3,x_4,x_5$ themselves? Or on your $a,b,c,d$? – Colin McLarty Mar 16 '14 at 19:52

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