Is it possible to construct an equilateral triangle with vertices on lattice points?

I think the answer is no, but how can I prove this?

I started with a triangle with coordinates $(0,0)$ $(a,b)$ and $(c,d)$. Equating the size of the 3 sides, I get


How should I continue?

I see there are solutions based on the fact that the angle between two edges can not be 60°. Is it possible to have a solution based on the fact that the length of the edges can not be the same?

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    $\begingroup$ This deals with a general version of your question. In particular, since $\tan\frac{\pi}{3}$ isn't rational, you can't have lattice points as the corners of an equilateral triangle. $\endgroup$ Commented Feb 3, 2012 at 15:02
  • $\begingroup$ possible duplicate of Which internal angles can a lattice polygon have? $\endgroup$
    – joriki
    Commented Feb 3, 2012 at 15:09
  • $\begingroup$ You've accepted an answer which I believe is incorrect. Please see my comment under the answer, and unaccept it in case you agree, as the checkmark will otherwise mislead others. $\endgroup$
    – joriki
    Commented Feb 3, 2012 at 15:23
  • 1
    $\begingroup$ $(a-c)^{2}+(b-d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}-2ab-2cd=2a^{2}+2b^{2}-2ab-2cd$ this implies $a^{2}+b^{2}=2ab+2cd$ $\endgroup$
    – wnvl
    Commented Feb 3, 2012 at 15:25
  • $\begingroup$ a) If you edit your post such that a correct earlier comment by someone else now appears wrong, please indicate this clearly (e.g. by adding "Edit" or the like). b) It's still wrong, since you've shown that $a^{2}+b^{2}=2ab+2cd$, whereas the post now says $a^{2}+b^{2}=2ac+2bd$. $\endgroup$
    – joriki
    Commented Feb 3, 2012 at 15:30

10 Answers 10


Let the vertices of our triangle be $(0,0)$, $(a,b)$, and $(c,d)$, where $a$, $b$, $c$, and $d$ are integers. If all edge lengths are the same, then $$a^2+b^2=c^2+d^2=(a-c)^2+(b-d)^2.$$ Minor manipulation turns this into $$a^2+b^2=c^2+d^2=2ac+2bd.$$

Now we use my favourite identity, which was known more than a millenium ago in India, and even earlier by Diophantus, and so has often been called the Fermat Identity: $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.\qquad\qquad(\ast)$$ This identity can be easily verified by expanding both sides, or more conceptually by noting that the norm of the product of two complex numbers is the product of the norms.

Let $N=a^2+b^2=c^2+d^2=**2(ac+bd)**$. Then $ac+bd=N/2$. The identity $(\ast)$ now gives $$N^2=\frac{N^2}{4}+(ad-bc)^2$$ or equivalently $$3N^2=4(ad-bc)^2.$$ This is impossible, since $3$ times the perfect square $N^2$ cannot be a square unless $N=0$, which gives a very tiny triangle.

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    $\begingroup$ A cute proof of the fermat identity: If $z = a+ib$ and $w = c + id$, then $ (zz') (ww') = (zw)(zw)'$ where $x'$ is the conjugate of $x$. $\endgroup$
    – Aryabhata
    Commented Feb 3, 2012 at 21:05

Solution 1 (by me): Assume WLOG that two of the points are $(0,0), (m,n), m,n \in \mathbb{Q}$. Then the third point is $(m/2 - n \sqrt{3} / 2, n/2 + m \sqrt{3}/2)$, which is not a rational point.

Solution 2 (by a friend): The determinant formula for area is rational, so if the all three points are rational points, then the area of the triangle is also rational, so whereas the area of an equilateral triangle with side length s is $\frac{s^2 \sqrt{3}}{4}$, which is irrational since $s^2$ is an integer.

Note that the above solutions both generalize from integer points to rational points.

You can also use Pick's theorem for integer points.

  • $\begingroup$ Very nice solution 2. $\endgroup$ Commented Apr 8 at 14:29

It is possible, but you need three dimensions in order to do it.

Consider $\bigtriangleup v_{1}v_{2}v_{3}$ with:




For $a,b\in{1,2,3}$, $a\neq b$, $d(v_{a},v_{b})=\sqrt{2}$, therefore the triangle is equilateral. It is not possible (as other answers indicate) to have an equilateral triangle with integer coordinates for the vertices in a two dimensional square lattice (a grid is just a 2d lattice).


My favorite proof: Pick's theorem says such a triangle would have rational area, but the area is a half integral multiple of $\sqrt{3}$.

  • $\begingroup$ +1. That was the first solution I thought of. $\endgroup$
    – Hans
    Commented Oct 6, 2019 at 21:17

I found the same problem on page 13 in Richard Courant's classic textbook Differential and Integral Calculus:

2.* In an ordinary system of rectangular co-ordinates, the points for which both co-ordinates are integers are called lattice points. Prove that a triangle whose vertices are lattice points cannot be equilateral

My proposed solution is to set the origin at a vertex. This can be done w.l.o.g. For every lattice point $P : (p,q)$ we get an equilateral triangle with vertex $P$ and a third vertex $X$ (we also get another triangle with $P$ and $X'$, see figure) where both vertices lie on the circumference of a circle with radius $r =\sqrt(p^2+q^2)$. The angle of the $OP$ edge is $\alpha$, and $\sin a = \frac q r, \ \cos \alpha = \frac p r$.

Figure: Two equilateral triangles for every lattice point $P$

We get the coordinates of $X$ on the circle (and in a similar way for $X'$):

$(x,y) = (r \cos(\alpha+\frac \pi 3), r \sin(\alpha + \frac \pi 3))$

The addition formulas for sine and cosine give,

$$(x,y) = \left(r \left[ \cos \alpha \cos \frac \pi 3 - \sin \alpha \sin \frac \pi 3 \right], r \left[ \sin \alpha \cos \frac \pi 3 + \cos \alpha \sin \frac \pi 3 \right] \right) = \left( r \left[ \frac p {2r} -\frac {q \sqrt 3} {2r} \right], r \left[ \frac q {2r} + \frac {p \sqrt 3} {2r} \right] \right) = \left( \frac p 2 - \frac{ \sqrt 3 q} 2, \frac q 2 + \frac{\sqrt 3 p} 2 \right)$$

If $ p,q \in \Bbb Z$ then $X$ is not a lattice point.


This is not an answer. It is really only to help a little with someone who is misreading an algebraic step.

We are given that $(a^2+b^2)=(c^2+d^2)=(a-c)^2+(b-d)^2$. I am sorry for the extra ()s but I worry that you are seeing = as +. so to be clear, I will write it out like this:




As you have written Jiten you know that $$(a−c)^2+(b−d)^2=(a^2+c^2)+(b^2+d^2)−2bd−2ac$$

Let us flip this around so that we have



I will rearrange symbols but change nothing:


Next we use the substitution:


We subtract.


And we arrive at



You can continue by: 2ac + 2bd is always even. So a and b are either both even, or both odd. Same for c and d. a, b, c and d could all be even, but this wouldn't be the minimal case.. If they were all even we could half a, b, c & d until at least 1 is odd. So at least one pair of (a,b) or (c,d) is odd. Let's say a and b are odd.. therefore a^2 + b^2 is not divisible by 4. This would make c and d odd as well. Similarly for c and d being odd. So a, b, c and d must all be odd. But that makes 2(ac + bd) divisible by 4, which is a contradiction.


Suppose that there were an example of positive edge-length. Consider an example of minimal edge-length.

Suppose that the vertices' $x$-coordinates all have the same parity. If their $y$-coordinates all have the same parity, then we could subtract 1 from any odd coordinates, then divide every coordinate by 2, yielding a smaller example, thus contradicting minimality.

So the $y$-coordinates must be of mixed parity. So label the vertices so that $A=(x_a, y_a), B=(x_b, y_b)$ have $y$-coordinates of the same parity, and let the other vertex be $C=(x_c, y_c)$. Then $|AB|^2=(x_a-x_b)^2+(y_a-y_b)^2$ is even, but $|AC|^2=(x_a-x_c)^2+(y_a-y_c)^2$ is odd.

Similarly if the $y$-coordinates all have the same parity.

If the $x$-coordinates are of mixed parity, as are the $y$-coordinates, then label the vertices so that $A=(x_a, y_a), B=(x_b, y_b)$ have $y$-coordinates of the same parity, and that $A$ and $C=(x_c, y_c)$ have $x$-coordinates of the same parity. Then $|AB|^2$ is odd, but $|BC|^2$ is even.

In each case, therefore, two sides must have squared lengths of different parity. Therefore $ABC$ cannot be equilateral.


If $P$ and $Q$ are lattice points, then $P$ rotated by 90 degrees about $Q$ is also a lattice point. If $R$ is also a lattice point, then so is $R$'s reflection across the line $PQ$.

So if these are lattice points:

enter image description here

Then so are these:

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And hence this one:

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So by rotational symmetry so are these:

enter image description here

But then we could get ever-smaller lattice equilateral triangles by infinite descent!


You can start like this. Without loss of generality, let the three points be $(0,0), (0,a)$ and $(b,a/2)$. You can do this because you can always rotate and translate the axis to get these points. Now for all the points to be integral, you need $a$ to be even. This is the first constraint. Secondly, from the basic trigonometry,

$$ \tan \theta = \frac{2b}{a} = \sqrt{3}. $$ From this, you get $b = a\sqrt{3}/2$, which is irrational.

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    $\begingroup$ You can't isometrically map any three lattice points to points of this form. The lengths of all sides may be irrational, whereas the length of the side from $(0,0)$ to $(0,a)$ is always an integer. $\endgroup$
    – joriki
    Commented Feb 3, 2012 at 15:17
  • $\begingroup$ Ohh yes! Thanks for pointing it out. $\endgroup$
    – Jalaj
    Commented Feb 3, 2012 at 22:43

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