Show that $\arctan(n)$ is irrational for all $n \in \mathbb{N}$ Question : Show that $\arctan(n)$ is irrational for all $n \in \mathbb{N}$.
Hint:
My solution doesn't use continued fraction.

I am interested in other possible proofs for this question.


Suppose that $\arctan n=r\in\Bbb Q$, where $n$ is a non-zero integer.  Then $r$ is not zero, so $2r$ is not zero, and
$$\cos2r=\frac{\cos^2r-\sin^2r}{\cos^2r+\sin^2r}
  =\frac{1-\tan^2r}{1+\tan^2r}=\frac{1-n^2}{1+n^2}$$
which is rational.  But this contradicts the result that the cosine of a non-zero rational number is irrational.

As for the proof of this result, it is usually done by taking an integral such as
$$\int_0^r f(x)\sin x\,dx\ ,$$
where $f(x)=x^n(a-bx)^{2n}(2a-bx)^n$ and $r=a/b$, and showing that if $n$ is large we get contradictory estimates for the integral.  See, for example, my lecture notes, starting at page 20.
