0

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]

KTF
  • 121
  • 1
    Welcome to Mathematics Stack Exchange. Did you mean for $n\color{red}\ge 1$? What are your thoughts? Look at $n=1$: $a_n=3$ and $b_n=1$ if the $4$ is left out – J. W. Tanner Jul 17 '20 at 01:11
  • 2
    You made a mistake when multiplying $(a_n+ib_n)(3+i)$ – J. W. Tanner Jul 17 '20 at 01:29
  • 1
    You corrected that mistake now – J. W. Tanner Jul 17 '20 at 01:44
  • Yes! Thank you for pointing that out! – KTF Jul 17 '20 at 01:47
  • 1
    "Show that for $n\ge1$" suggests mathematical induction; are you familiar with that? – J. W. Tanner Jul 17 '20 at 01:48
  • Start by proving n = 1 then prove for n + 1, right? So for n = 1, a = 3 and b = 1. For n = 2 we have a = 8 and b = 6. So I get that it's some multiple of 5 plus 3 and 1 respectively, but I'm not sure how to prove it – KTF Jul 17 '20 at 01:58
  • 1
    If $a_n\equiv3$ and $b_n\equiv1$, then $a_{n+1}=3a_n-b_n\equiv3\times3-1=8\equiv3\pmod5$ – J. W. Tanner Jul 17 '20 at 02:00
  • Okay! So we can just use {a_1} and {b_1} as {a_n} and {b_n}? If you don't mind my asking, is there a particular reason this is allowed? – KTF Jul 17 '20 at 02:04
  • 1
    With mathematical induction, you show the base case $(n=1)$, and then you show that, assuming it's true for $a_n$, it's true for $a_{n+1}$ – J. W. Tanner Jul 17 '20 at 02:16
  • 1
    In a proof by induction, we show that the proposition is true in some base case (n=1 here). Then show that when the proposition hold hold for some non-specific n, it also holds for n+1. This implies it will hold for all natural numbers. i.e. it holds for 1 implies it holds for 2 which in turn implies it holds for 3, etc. – Doug M Jul 17 '20 at 02:17

3 Answers3

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.

hbghlyj
  • 2,115
Doug M
  • 57,877
1

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

lhf
  • 216,483
0

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

lhf
  • 216,483