need help with this q (1 Viewer)

[lmao1010]

any help appreciated

 

[jimmysmith560]

Would this help for question 19?

 

[5uckerberg]

Do you need help for Q17?

 

[lmao1010]

Do you need help for Q17?

Yep

 

[username_2]

View attachment 34847
View attachment 34848
any help appreciated

Alright so a fairly straightforward apporach to this question:

So what I have written below is the solution and no hints so be wary.

17)
a) Remember show not equals prove, hence we just need to show that LHS = RHS

LHS = x^n - 1
RHS =
(x - 1) (x^n-1 + x^n-2 + ... + x^2 + x + 1)
= [First expand with x]
x^n + x^n-1 + x^n-2 + ... + x^2 + x
[Now expand with -1]
- x^n-1 - x^n-2 - ... - x^2 - x - 1

= [After cancelling out] = x^n - 1 = LHS

Hence, LHS = RHS, and 17) a) is true

b) Using a) we can replace x = 7, resulting in: 7^n - 1 = (7-1) (7^n-1 + .. + 1)

As n is always an integer, (7^n-1 + .. + 1) is always going to be an integer whole value, meaning that the (7^n - 1) term has two factors 7-1 = 6 and the other integer.

Hene, as 6 is a factor of the original term, 7^n - 1 is divisible by 6

For ii)
For any value of a^n - 1 to be prime, a must be an integer and must be written in the form:

a^n - 1 = (a - 1) (a^n-1 + a^n-2 + ... + a^2 + a + 1) = 1 x (a^n - 1)

Here there are two possibilities, either (a - 1) = 1 or (a^n-1 + a^n-2 + ... + a^2 + a + 1) = 1

For the first possibility, it is very obvious that this is only possible when a = 2 by simple rearrangement.
For the second possibility, a^n-1 + a^n-2 + ... + a^2 + a = 0 and for this to be possible, a = 0

But it is well know that if all of the values are 0, then a^n - 1 = -1, which is not prime, meaning that the second possibility leads to a contradiction.

Hence, it is true that only a = 2 is a solution if it is known that a^n - 1 is prime.

Hope this helps.