Why does that fact that $a_n \equiv 3 \pmod 5$ and $b_n\equiv 1 \pmod 5$ imply $1/\pi(\arctan(1/3))$ is irrational

Question

So since I've started this multi-part question I've learned:

$(3+i)^n = a_n+ib_n$

$a_{n+1} = 3a_n-b_n$

$b_{n+1} = 3b_n+a_n$

$a_n \equiv 3 \pmod 5$

$b_n \equiv 1 \pmod 5$

Now I am asked why the fact that, for $n\geq 1$, $a_n \equiv 3 \pmod 5$ and $b_n \equiv 1 \pmod 5$ implies that $\frac{1}{\pi} \arctan(\frac{1}{3})$ is irrational.

I notice the similarities between $\frac{b_n}{a_n}$ and $\frac{1}{3}$, but I honestly have no clue how to go about this.

Answer 1

Let $z=3+i$. Then $\arg z = \arctan ⅓ =: t$.

Now, since $b_n \neq 0$, we have that $nt = m\pi$ is impossible for any integer $n,m$.

(Note: $nt=\arg z^n$ comes from repeatedly taking powers of $z$. Suppose it can equal $m\pi$ for some integer $m$, then we must have $b_n=0$ for some $n$)

Thus, $t/ \pi = m/n$ has no solutions over the integers, so this number is irrational.