Find a counterexample to the statement 4^n + 1 is divisible by 5...how do i show my work for this problem? The example in my book is too confusing it doesnt really go into detail but just shows a table.

What happens if yo put n = 1?

I get 4^1+1=5 divided by 5 =1?

NVM i figured it out =)

n=2

4 = -1 [5]

4^n = (-1)^n [5]
4^n + 1 = (-1)^n + 1 [5]

so if n=2k+1 4^n + 1 is divisible by 5

To find a counterexample for the statement that 4^n + 1 is divisible by 5, we need to show a specific value for 'n' that makes the statement false.

To begin, let's examine the expression 4^n + 1. We want to test if this expression can be divided evenly by 5 for any value of 'n'. Since the exponent 'n' can be any positive integer, we can start by trying a few different values to see if any of them result in a non-divisible outcome.

Let's start with the simplest possible value for 'n', which is 1. In this case, we get 4^1 + 1 = 5. Since 5 is divisible by 5, it does satisfy the statement.

Now, let's try the next value for 'n', which is 2. Plugging in 'n = 2' into the expression yields 4^2 + 1 = 16 + 1 = 17. Since 17 is not divisible by 5, we have found a counterexample.

Therefore, 'n = 2' is a counterexample to the statement that 4^n + 1 is divisible by 5.

To show your work for this problem, you can lay it out as follows:

Statement: 4^n + 1 is divisible by 5.
Counterexample: Let's find a value of 'n' that makes the statement false.

Attempt 1: For n = 1,
4^1 + 1 = 5, which is divisible by 5.

Attempt 2: For n = 2,
4^2 + 1 = 16 + 1 = 17, which is not divisible by 5.

Therefore, n = 2 is a counterexample, and it demonstrates that the original statement is false.