For $n \in \Bbb{Z}_{>0}$, let $(3+i)^n = a_n+ib_n$, where $a_n, b_n \in \Bbb{Z}$.

Question

Find expressions for $a_{n+1}$ and $b_{n+1}$ as linear combinations of $a_n$ and $b_n$ with coefficients independent of n. With some of your comments, I see $a_{n+1} +ib_{n+1} = (a_n+ib_n)(3+i) = 3a_n + ia_n + i3b_n-b_n$. So the imaginary parts have to be equal which means that $b_{n+1} = a_n +3b_n$ and the real parts have to be equal so $a_{n+1} = 3a_n - b_n$, right? So that's that question I believe

Show that for $n\geq 1, a_n \equiv3\pmod 5$ and $b_n \equiv 1\pmod 5$ Here we know that if n = 1, $(3+i) = a_1 + ib_1$ so $a_1 = 3$ and $b_1 = 1$ which means that $a_1 \equiv 3(mod 5)$ and $b_1 \equiv 1 (mod 5)$. We can use these as our base case for $a_n$ and $b_n$ and see that $a_{n+1} = 3*3-1 = 8 \equiv 3(mod 5)$ and $b_{n+1} = 3*1+3 = 6 \equiv 1(mod 5)$. [Thank you to J.W. Tanner for your help with this]

Answer 1

Hint: $a_{n+1}+i b_{n+1} = (a_n+ib_n)(3+i)$.

Answer 2

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

Proposition: $a_n \equiv 3 \pmod 5, b_n \equiv 1 \pmod 5$

Proof by induction

Base case: $n = 1$
$a_n = 3, b =1$

Suppose the proposition is true.

$(3+i)^{n+1} = (a_n+b_ni)(3+i) = 3a_n + a_n i - b_n +3b_ni$

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

By the inductive hypothesis we can conclude:
$3\cdot 3 - 1 \equiv 3 \pmod 5\\ 3\cdot 1 + 3 \equiv 1 \pmod 5$

As for your next question...

By De Moivre:
$(3+i)^{n} = (\sqrt{10})^n(\cos (n\arctan \frac 13) + i\sin (n\arctan \frac 13))$

$a_n = \sqrt{10}\cos (n\arctan \frac 13)$

We know that $a_n$ is an integer.

For all $n,$ either $\cos (n\arctan \frac 13)$ is rational, or $\sqrt{10}\cos (n\arctan \frac 13)$ is rational.

As for $\frac 1{\pi}\arctan \frac 13$... I am not sure what to say.

Answer 3

Hint: $(3+i)^2 \equiv 3+i \bmod 5$