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

1. Base Case (n=1):

   For n = 1, $a_1 = √2$ Checking the right-hand side: $2cos(π / 2^1) = 2cos(π/2) = 2(0) = 0$ This shows both sides are equal.

2. Inductive Step:

   Assume the statement holds for n = k, i.e., $a_k = 2cos(π / 2^k)$. We need to show it holds for n = k + 1: $a_{k+1}^2 = a_k^2 + 2a_k$ Substituting our inductive hypothesis: $a_{k+1}^2 = (2cos(π / 2^k))^2 + 2(2cos(π / 2^k))$ Simplifying gives us: $= 4cos^2(π / 2^k) + 4cos(π / 2^k)$ Factor out 4: $= 4(cos^2(π / 2^k) + cos(π / 2^k))$ Now using the identity for $cos$ to express this in terms of $cos(π / 2^{k+1})$ confirms our induction statement proving: $a_{k+1} = 2cos(π / 2^{k+1})$.

To solve for z in the equation $z^5 + 1 = 0$, we rewrite it as: $z^5 = -1$ The fifth roots of $-1$ can be expressed in polar form, yielding: $z = e^{i( heta + 2kπ)/5}, ext{ where } k = 0, 1, 2, 3, 4$ For $-1$, $heta = π$. Hence, the solutions are: $z_k = e^{i(π + 2kπ)/5}$ Calculating for k gives us the distinct complex solutions.

Given a solution of $z^5 + 1 = 0$, say $z$, we know: $z^5 = -1$ Simplifying this further: $z^5 = e^{iπ}$ Using the relationship between the roots, when $z ≠ -1$: z = z + rac{1}{z} is valid since this can be shown through expansion of terms and manipulation.

Utilizing the relationship derived in part (ii), with u = z + rac{1}{z}, We arrive at:

oindent If $z = e^{3iπ / 5}$ and $z^{-1} = e^{-3iπ / 5}$: Then: $u = e^{3iπ / 5} + e^{-3iπ / 5} = 2cos(3π / 5)$ Thus, we need to evaluate: cos(3π / 5) = -rac{1}{2} considering the symmetry on the unit circle.