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This is a follow-up question to this one. I asked it there first but moved it here following the advice from Cam McLeman.

I tried to prove that $(\mathbb P:\mathbb Q)=\aleph_0$ and I think I succeeded: it's clear that the square of every natural number is in $\mathbb P$. Indeed, $\sqrt n=(\underbrace{1^2+1^2+\ldots+1^2}_n)^{1/2}\in\mathbb P.$ In this question I learned how to prove that the squares of primes are linearly independent over $\mathbb Q.$ Thus $\{\sqrt p\,|\,p \text{ is prime}\}\subset \mathbb P$ is an infinite linearly independent set.

Can we construct a $\mathbb Q$-basis of $\mathbb P?$ We have for $p_1,p_2,\ldots,p_m,q_1,q_2,\ldots,q_n\in \mathbb Q$

\frac 1 {\sqrt{p_1^2+p_2^2+\ldots+p_m^2}}=\frac{1}{p_1^2+p_2^2+\ldots+p_m^2}\cdot\sqrt{p_1^2+p_2^2+\ldots+p_m^2} and \sqrt{p_1^2+p_2^2+\ldots+p_m^2}\cdot \sqrt{q_1^2+q_2^2+\ldots+q_n^2}=\sqrt{\sum_{1\leq i\leq m,1\leq j\leq n}(p_iq_j)^2}.

So I think every number in $\mathbb P$ should be possible to write as a linear combination of numbers of the form $\sqrt{p_1^2+p_2^2+\ldots+p_m^2}.$ But the set of such numbers cannot possibly be linearly indepentent over $\mathbb Q.$ (Although I think it's a step in the right direction, isn't it?)

EDIT I'll try to clarify why I think the above equations are relevant. I think I know that every number in $\mathbb P$ is a result of iterating field operations on the set of numbers of the form $\sqrt{p_1^2+\ldots +p_n^2}.$ The above equations tell me that I don't need muliplication and multiplicative inversion in this procedure. So every number in $\mathbb P$ is a result of iterating addition and additive inversion on the set of numbers of the form $\sqrt{p_1^2+\ldots +p_n^2}.$ These two happen to be vector space operations. So this set generates the whole $\mathbb P$ as a vector space.

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    @ymar: I did it once long ago, to win an argument about whether Cantor's proof of the existence of transcendentals is constructive (it is). Do not know a standard source for the details.2012-03-05

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