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I am looking for a solution for a conjecture as follows.

In Cartesian plane, no exist an equilateral triangle such that three vertices are integer numbers.

I hope that you like the question and let me a answer.

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    $\begingroup$ Are you measuring angles in degrees or radians? If degrees, take an equilateral triangle with side lengths $1$. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 19, 2016 at 15:38
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    $\begingroup$ Also the sum of angles must be equal to $\pi$ so of course no all three angles can be rational. $\endgroup$
    – Loïc Teyssier
    Commented Mar 19, 2016 at 15:39
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    $\begingroup$ Nice question. Wrong forum. It might work for math.stackexchange. (Also, it has been proved that equilateral triangles embed in Z^3 and not Z^2.) Gerhard "This Is Not Math.StackExchange Forum" Paseman, 2016.03.19. $\endgroup$
    – Gerhard Paseman
    Commented Mar 19, 2016 at 15:51
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    $\begingroup$ This has many nice proofs and generalizations. For example, compare the area formula for equilateral triangle with side $a$ $\sqrt{3}/4 a^2$ and the fact that double area must be integer. $\endgroup$
    – Fedor Petrov
    Commented Mar 19, 2016 at 15:52
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    $\begingroup$ My favorite proof of this is to observe that if you can do it with an equilateral triangle, then by a few reflections you can do it with a regular hexagon, and then there's a nice visual proof that a regular hexagon cannot be put on the integer lattice: mathoverflow.net/a/25305/5701 $\endgroup$
    – Vaughn Climenhaga
    Commented Mar 19, 2016 at 16:19

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Let A,B,C be the vertices.Use the fact that the area is $E=\frac{1}{2}\cdot |det(\vec{AB},\vec{AC})|$
Since $det(\vec{AB},\vec{AC})$ is an integer and the area must be of the form
$AB^2\cdot \frac{\sqrt3}{4}$ which is not an integer you can see the contradiction

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