# Equilateral triangle with vertices whose coordinates on the Cartesian plane are integers. Does such a triangle exist? [duplicate]

Can you build an equilateral triangle on a Cartesian plane whose vertices only have integer values as their coordinates?

Looking at the simplest example, i.e. a triangle with vertices (0,0), (1,0) and (0,1), this doesn't look possible as there is a square root involved, but can you build one such triangle at all?

We can assume without restricting generality, that one of the triangle's side is on the $\;x -$ axis , so its vertices can be put as $\;(0,0),\,(a,0),\,\left(\frac a2,\,b\right)\;$ , since the middle point of the horizontal base is on the perpendicular to it through the upper vertex (in an equilateral triangle all opposite side bisector, upper angle's bisector and height to opposite side are texactly the same). From here that $\;a\;$ must be an even number, but also the lengths: the length of the base must equal the length of any other side, so

$$a=\sqrt{\frac{a^2}4+b^2}\iff 2a=a^2+4b^2\iff a^2-2a+4b^2=0$$

The above is an integer quadratic and its discriminant, which must be non-negative (otherwise the above is impossible), is

$$\Delta:=4-16b^2=4(1-4b^2)\ge0\iff 4b^2\le1$$

and this last equality is impossible for integer $\;b\neq0\;$, and this is already a contradiction since, by construction, clearly $\;b\neq0\;$ .

Other wayL since the upper vertex has integer entries, if we take half a triangle we get by Pythagoras that

$$\left((b,0)-(0,0)\right)^2+b^2=\frac{a^2}4+b^2\iff b^2=\frac{a^2}4\iff (2b-a)(2b+a)=0$$

If we decide to construct our triangle in the first quadrant , the above is possible iff $\;a=2b\;$, but again equating base and the rightmost side we get

$$2b=\sqrt{2b^2}\stackrel{\text{Because}\;b\neq0}\iff 2=\sqrt2$$

and ,again, impossible.

• Very nice, thank you sir – user1301428 Mar 16 '16 at 8:07