Question

Question 9 please both a and b

Answer 1

$a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + ... + b^{k-1})$

a. Let's set a=3, b=1

$3^k - 1^k = (3 - 1)(\textrm{some integer we'll call } m)$

$3^k = 1 + 2m$

In other words $3^k$ is odd, i.e. $3^k$ has remainder 1 when divided by 2. $\quad\checkmark$

b. Let

$k=2n, a=3, b=1$

$a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + ... + b^{k-1})$

$3^{2n} - 1^{2n} = (3 - 1 )(3^{2n-1} + 3^{2n-2} + ... + 1)$

There are 2n terms in the last factor.  All the terms of the form

$3^{2n-k}$

are odd. So when we add up an even number of them, we get an even number, call it 2m.

$3^{2n} - 1 = 2(2m)$

$3^{2n} - 1 = 4m$

We've shown $3^{2n}-1$ is divisible by four.  $\quad \checkmark$

Merry Christmas!