- #1
cragar
- 2,552
- 3
Homework Statement
Prove [itex] 4^{2n}-1 [/itex] is divisible by 15 for all positive integers
The Attempt at a Solution
Ok so I factor it into this
[itex] (2^n+1)(2^n-1)(4^n+1) [/itex]
I know how to get the factor of 3 because we know
2^n is even so either 2^n+1 or 2^n-1 is divisible by 3 because we have 3 consecutive integers so at least one of them needs to be divisible by 3.
But I am not sure how to get the factor of 5.
I thought about squaring some of their components and looking how close they
are to 4^n+1.