# Constructible integral degree angles.
For which integer $x$ is the angle $x^{\circ}$ constructible?
It is well-known that equilateral triangles and regular pentagons are constructible. This means the angles $60\degree$ and $72\degree$ are constructible.
Whence we can construct their difference $72\degree - 60\degree = 12\degree$.
By repeated bisection, we can construct $12\degree / 2 = 6\degree$ and $6\degree / 2 = 3\degree$.
This shows we can construct $3\degree$ and any integral multiples of $3\degree$, $3n\degree$.
Now we show that we cannot construct other integer degrees that are not multiples of $3$. Suppose to the contrary that some $y\degree$ is constructible, $y$ some integer but $y$ is not a multiple of $3$. Then by repeated subtracting $3$, we can construct either $2\degree$ or $1\degree$ using $y\degree$. And if we can construct $2\degree$, then with $3\degree$ we can further construct $1\degree$. So if we can construct an integer degree angle that is not a multiple of $3\degree$, then we can construct $1\degree$. But this implies we can construct any integer multiple of $1\degree$! But it is well known that we cannot construct $20\degree$! Contradiction.
Hence we have
> Theorem.
> If $x$ is an integer, then $x\degree$ is constructible if and only if $3 \mid x$.
Neat!