Does closure under Euclidean distance imply closure under square roots (in this set)? Suppose you start with the number 1 (and 0 because why not) and the set is closed so that you may add, subtract (smaller from bigger), multiply, and divide (except 0) any elements of the set. It is easy to see all non-negative rational numbers are elements of this set. If we now also allowed for Euclidean distance as well, what would this set look like? That is, $a,b\in S \Longrightarrow \sqrt{a^2+b^2}\in S.$ For starters, all square roots of rational numbers will be in the set (simple to prove for integers, and the rationals follow by closure under division).
However, I wonder if I can take the square root of any element of the set. For example, are the higher (even) roots of rationals included? What about the root of one plus some other root? Basically, does $x \in S \Longrightarrow \sqrt{x} \in S$?
It has proven difficult to give this set an explicit representation. That's why my attempts to solve equations or find a counter-example have ended without success so far. If anyone could help, I would gladly appreciate it!
EDIT: While this apparently is a Pythagorean field, the reverse does not seem to be true. For example, $\sqrt{\frac{1}{2}}$ does not have a representation under Pythagorean fields at all. So, the question whether Pythagorean fields are Euclidean Fields seems easily answered by the counterexample $\sqrt{\sqrt{2}+1}$. However, this does not seem applicable for the case $a\sqrt{1+b^2}$ which it seems I must pay attention to.
 A: An ordered field $P$  such that, given any element $x \in P,$   we also have $ \sqrt{1+x^2} \in P.$  Such a field is called Pythagorean. This is your example:  given field elements $a,b$ with nonzero $a >0,$   we see that  $\frac{b}{a}  \in P, \; \; $ and $ \sqrt{1 +  \frac{b^2}{a^2}  }  \in P . \; \; $   Then $$   \sqrt{a^2+b^2} \;  = \; \; \; \; a \; \sqrt{1 +  \frac{b^2}{a^2}  }  \;  \in \; P  \; \; .$$
A: After quite a bit of research, this is difficult to prove properly! It relies on finding some kind of invariant that stays the same regardless which of these operations is applied. Here are some ideas from Auckly, D. (1995). Totally Real Origami and Impossible Paper Folding. The American Mathematical Monthly.

*

*Algebraic numbers $\alpha$ are the zeros of polynomials with rational coefficients. Moreover, there is a unique minimal (irreducible) polynomial in $\mathbb{Q}[x]$ for which $\alpha$ is a zero. The other roots of this polynomial are the conjugates of $\alpha$. We call algebraic numbers totally real iff all its conjugates are real.

*To show: The sum, difference, product, quotient, Euclidean distance of totally real numbers is totally real.

*This is done using symmetric polynomials. They remain unchanged when swapping terms (such as $x^2+y^2$ but not $x^2-y^2$). Also, every symmetric polynomial is a linear combination of products of three so-called elementary symmetric polynomials: $s=\sum_{i=1}^{n}x_i$, $p=\prod_{i=1}^{n}x_i$ and $s'$, the sum of all products of $k$ distinct $x_i$.

*We eventually find $\sqrt{x}\in S \Longrightarrow \sqrt{x}$ is totally real.

*Then not all conjugates, for example, of $\sqrt{1+\sqrt{2}}$ are real. Hence, $\sqrt{1+\sqrt{2}} \not\in S$.

The quoted source has the most detailed proof I could find online. Thanks again for everyone's help!
