Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

We start with the base case where n = 1:

$a_1 = \sqrt{2} = 2cos\left( \frac{1\pi}{2*1+1} \right) = 2cos\left( \frac{\pi}{3} \right).$

For the induction step, assume that the formula holds for n = k:

$a_k = 2cos\left( \frac{k\pi}{2k+1} \right).$

Now we need to prove it for n = k + 1:

Using the relation $a_{n+1}^2 = 2 + a_n$:

$a_{k+1}^2 = 2 + 2cos\left( \frac{k\pi}{2k+1} \right).$

Using the cosine addition formula:

$= 2 + 2\cdot 2cos^2\left( \frac{k\pi}{2k+1} \right) - 1 = 2\cdots cos\left( \frac{(k+1)\pi}{2(k+1)+1} \right),$

Thus, the proof holds for k + 1 and by mathematical induction, it holds for all integers n ≥ 1.

The solutions to the equation $z^5 + 1 = 0$ can be found by rewriting it as:

$z^5 = -1.$

In complex form, -1 can be written as:

$-1 = e^{i\pi + 2k\pi i}, k \in \mathbb{Z}.$

Thus,

$z = e^{\frac{i(\pi + 2k\pi)}{5}} = e^{\frac{i\pi}{5}}$ for k = 0, 1, 2, 3, 4, so we have 5 solutions:

Assuming $z^{5} = -1$, we can express $u = z + \frac{1}{z}$. Squaring both sides gives:

$u^2 = z^2 + 2 + \frac{1}{z^2}.$

Using the fact $z^5 + 1 = 0$, thus:

$z^5 = -z^{-5},$

which leads to:

$z^2 + \frac{1}{z^2} = 1.$

Hence, we have shown that $u$ satisfies the relation.

By part (i), we have found the solutions:

$e^{\frac{3\pi}{5} i}, e^{\frac{3\pi}{5} i} = 2cos\left( \frac{3\pi}{5} \right).$

Thus, the exact value is:

$\cos\left( \frac{3\pi}{5} \right) = -\frac{1 + \sqrt{5}}{4}.$