# Proving co-ordinates of an equilateral triangle are integers in a plane

Consider the 2D plane $P$ in $\Bbb{R}^3$ defined by $$P=\{x \in \Bbb{R}^3 \mid x_1+x_2+x_3=0\}.$$ Let $a$, $b$, $c$ be the vertices of an arbitrary equilateral triangle in $P$ such that all the co-ordinates of both $a$ and $b$ are integers. Prove that the coordinates of $c$ are also integers.

I can prove it for individual cases pretty easily by assigning values to $a$ and $b$ and then calculating their midpoint, the length of the side and solving the resulting simultaneous equations for $c$, but I can't work out how to get started on the general case.

$$c=\frac{a+b}2\pm\frac n2\times(a-b)=\frac{a+b\pm n\times(a-b)}2$$
(where $n=(1,1,1)$ is orthogonal to the plane), the parity of each coordinate of the numerator is the sum of the parities of all coordinates of $a$ and $b$. Since these coordinates add up to $0$, the coordinates of the numerator are even.
• @Nemo: The length of $a-b$ is the side length. Thus to get the altitude $c-(a+b)/2$, we need to take the cross product of $a-b$ with a vector that's orthogonal to the plane and has length $\sqrt3/2$, the ratio of the altitude to the side length. The vector $n/2$ has that length. – joriki Sep 3 '15 at 21:56
• @Narasimham: It's not clear to me which generalization you have in mind -- do you want to deduce integer coordinates for $n-2$ points from those of $2$ points? Or for $1$ point from those of $n-1$ points? – joriki Sep 4 '15 at 8:46
• @Narasimham: I assume that what you posted as an answer was meant to be a comment in response to my above comment. For a square, integer coordinates for $2$ points do not imply integer coordinates for the other $2$ points -- consider e.g. the square spanned by $(0,0,0)$, $(1,-1,0)$ and $(1,1,-2)/\sqrt3$. Integer coordinates for $3$ points do imply integer coordinates for the $4$-th point, as the $4$-th point can be obtained as a linear combintion of the other $3$ with integer coefficients. – joriki Sep 4 '15 at 11:02