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When a given author has developed the basics of the geometry of numbers, he/she usually proceeds to exemplify the theory by proving Lagrange's theorem on the representability of the natural numbers as a sum of four perfect squares. The proof is structured more or less as follows:

1) It is noted that it suffices to prove the result for any odd prime number $p$ (because of the well-known identity for the product of two numbers that are the sum of four squares).

2) If $p$ is a prime number then, appealing to the Schubfachprinzip, he/she easily establishes the existence of $a,b \in \mathbb{Z}$ such that $a^{2}+b^{2}+1\equiv 0 \pmod{p}$.

3) Minkowski's theorem is applied then to the convex body $$\mathcal{K} := \{(w,x, y,z) \in \mathbb{R}^{4}\colon w^{2}+x^{2}+y^{2}+z^{2}<2p\}$$ and the lattice $$\Lambda := \langle \mathbf{a}_{1}, \mathbf{a}_{2}, \mathbf{a}_{3},\mathbf{a}_{4}\rangle \subseteq \mathbb{R}^{4}$$ where $\mathbf{a}_{1}=(p,0,0,0)$, $\mathbf{a}_{2}=(0,p,0,0), \mathbf{a}_{3}=(a,b,1,0)$, and $\mathbf{a}_{4}=(-b,a,0,1)$.

There is no denying this proof is pretty simple, aight? (IMHO, it is even simpler than the one in which one ends up resorting to descent.) On this night's Plutonian shore, my question has to do with the choice of $\mathbf{a}_{4}$. I believe I can see why it is that the "first" three elements of the above $\mathbb{Z}$-basis of $\Lambda$ were chosen that way (we need the absolute value of the determinant of the matrix of the basis to be equal to $p^{2}$; it would be just wonderful if the said matrix were upper or lower triangular, right?, etc.).

Do you know why it is that $\mathbf{a}_{4}$ is not simply taken as $(a,b,0,1)$ or $(b,a,0,1)$? Is is really necessary to have that $-b$ or swap the positions of $a$ and $b$ while we are at it? I think the question is a legitimate one; in the expositions of this proof upon which I have laid my eyes, one of the generators of the lattice $\Lambda$ usually resembles that $(-b,a,0,1)$. What's going on here?

Thanks in advance for your replies.

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    $\begingroup$ What do you find when you go through the proof with $\mathbf a_4$ equal to $(a,b,0,1)$ or $(b,a,0,1)$? $\endgroup$
    – KCd
    Feb 16, 2018 at 7:51
  • $\begingroup$ Duh! What a dunce I am. $\endgroup$
    – Jamai-Con
    Feb 16, 2018 at 8:01

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You need the dot products of the vectors ${\bf a}_i$ to be divisible by $p$. The dot product of $(-b,a,0,1)$ with $(a,b,1,0)$ is $0$ which is good, but that of $(b,a,0,1)$ with $(a,b,1,0)$ is $2ab$, unlikely to be divisible by $p$.

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