# Is surface $x^2+z^2=2\cdot y^2$ something of a Möbius strip?

This question is naive. My association with Möbius strip comes from not being able to smoothly extract positive solutions of the diophantine equation

$$x^2+z^2=2\cdot y^2$$

I got a parametrization (must be classical):

• $\quad x\ :=\ \alpha^2 + 2\alpha\cdot\beta - \beta^2$

• $\quad y\ :=\ \alpha^2 + \beta^2$

• $\quad z\ :=\ \beta^2 + 2\alpha\cdot\beta - \alpha^2$

(see that   $z^2-y^2=y^2-x^2$),   and I considered also the other three related parametrizations, obtained by replacing one or the both   $x\ z$   by   $-x\ \ -\!z$   respectively. I still don't seem to parametrize the positive solutions alone. It feels that the positive solutions flow seamlessly into the negative solutions (or mixed solutions). I'd appreciate some expert comments about this situation to educate me, please.

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The positive solutions are just the ones where $\sqrt{2}-1 < \alpha/\beta < \sqrt{2}+1$. They therefore form a triangular slice of the solutions overall. In general, one should never expect to be able to extract positive solutions, except by the use of inequalities - since being positive is, after all, an inequality. – Will Sawin Jun 30 '13 at 23:04
@Will Savin, thank you. (I'll still look at this surface, some things about it still feel to me interestingly odd). – Włodzimierz Holsztyński Jun 30 '13 at 23:59
There should be no trouble isolating the positive solutions by the method here: en.wikipedia.org/wiki/Tree_of_Pythagorean_triples – Will Jagy Jul 1 '13 at 0:59
Also, if you put positive (primitive) Pythagorean triples $a^2 + b^2 = c^2$ with $a$ odd, then $$(a,b,c) \mapsto (|a-b|, a+b, c)$$ is a bijection from the positive Pythagorean triples to your triples, thus giving you a perfectly good tree. – Will Jagy Jul 1 '13 at 3:38
-1 What is your question? – S. Carnahan Aug 6 '14 at 7:57

Ignoring the question itself, I was curious to see what the surface looks like. So here is a plot within $[\pm 1]^3$:
instead $\arctan \sqrt 2$ – Will Jagy Jul 1 '13 at 0:20