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

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 $$y$$.

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 $$|a|_2\ne |b|_2$$.

• Are you Hagen, or his "Doppelgänger"? – Dietrich Burde Dec 19 '18 at 9:44
• @i am his Doppelgänger who forgot to take his correct logon cedentials for vacatopn :) – Hagen von Eitzen Dec 19 '18 at 9:48

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.

• Ah good, so I wasn't doing anything wrong by failing to derive it from the metric space rules alone. – samerivertwice Dec 19 '18 at 9:50