[tex]a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + ... + b^{k-1})[/tex]
a. Let's set a=3, b=1
[tex]3^k - 1^k = (3 - 1)(\textrm{some integer we'll call } m)[/tex]
[tex]3^k = 1 + 2m[/tex]
In other words [tex]3^k[/tex] is odd, i.e. [tex]3^k[/tex] has remainder 1 when divided by 2. [tex]\quad\checkmark[/tex]
b. Let
[tex]k=2n, a=3, b=1[/tex]
[tex]a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + ... + b^{k-1})[/tex]
[tex]3^{2n} - 1^{2n} = (3 - 1 )(3^{2n-1} + 3^{2n-2} + ... + 1)[/tex]
There are 2n terms in the last factor. All the terms of the form
[tex]3^{2n-k}[/tex]
are odd. So when we add up an even number of them, we get an even number, call it 2m.
[tex]3^{2n} - 1 = 2(2m)[/tex]
[tex]3^{2n} - 1 = 4m[/tex]
We've shown [tex]3^{2n}-1[/tex] is divisible by four. [tex]\quad \checkmark[/tex]
Merry Christmas!
