Are there any equilateral triangles in a p-adic field?

Question

I'm struggling to find an equilateral triangle in $\Bbb Z_2$ so I wondered if there aren't any.

Suppose the triangle $\{x,y,z\}$

If this is to be equilateral then $\lvert x-y\rvert_p=\lvert z-y\rvert_p=\lvert x-z\rvert_p=r$

The strong triangle inequality quickly gives us that every triangle is isosceles.

When I try to choose a 3rd point such that the triangle is equilateral I always seem to be frustrated, but I can't put my finger on the algebraic proof that to do so is impossible. It seems I should think about the balls radius $r$ around $x,y$ then show no $z$ exists in both balls which is not closer than $r$ to either $x$ or $y$.

Answer 1

By translation, we may assume wlog that $x=0$, and by scaling we may asssume that one of $y,z$ is an odd integer. Then for an equilateral triangle, we Need that both $y,z$ are odd - but that make $y.z$ even.

Put differently, we have $|a+b|_2\le \max\{|a|_2,|b|_2\}$ with equality iff $|a|_2\ne |b|_2$.

Answer 2

As Hagen von Eitzen pointed out, this cannot be done in $\Bbb{Z}_2$. If you are given three integers some two of them will be congruent modulo two (and the same also holds in the 2-adic completion). Basically the pigeonhole principle in action.

Of course, if $p>2$ then the choices $x=0,y=1,z=2$ will yield an equilateral $p$-adic triangle.

With $p=2$ you need to go to an unramified extension to do the same. Say, instead of $\Bbb{Z}_2$ let's look at $\Bbb{Z}_2[\omega]$ with $\omega$ a primitive third root of unity (and a solution of $\omega^2+\omega+1=0$). Then the (extensions of) 2-adic distances between all four of $0,1,\omega,\omega^2$ are all equal to one.