Prove that 4^(n+1) + 5 is divisible by 3

[XJellieBX]

Prove that 4^(n+1) + 5 is divisible by 3 for all non-negative integers n.

Here's what I did:
The Base Case, when n=0
4^(0+1) + 5 = 4 + 5 = 9 --> divisible by 3

Assume that 4^(k+1) + 5 is also divisible by 3 for all n=0,1,2,...k.
Then it must also be true for k+1 and
4^(k+1+1) + 5 = 4 x 4^(k+1) + 5

My solution doesn't seem to be complete. Is there something wrong or missing?

 

[HallsofIvy]

Prove that 4^(n+1) + 5 is divisible by 3 for all non-negative integers n.

Here's what I did:
The Base Case, when n=0
4^(0+1) + 5 = 4 + 5 = 9 --> divisible by 3

Assume that 4^(k+1) + 5 is also divisible by 3 for all n=0,1,2,...k.
Then it must also be true for k+1 and
4^(k+1+1) + 5 = 4 x 4^(k+1) + 5

My solution doesn't seem to be complete. Is there something wrong or missing?

Well, yes, it is surely not obvious, yet, that your last formula is divisible by 3: which is the whole point.

How about writing 4 x 4^(k+1) as 3 x 4^(k+1)+ [4^(k+1)] so that 4 x 4^(k+1)+ 5 is equal to 3 x 4^(k+1)+ [4^(k+1)+ 5]? Can you see now that both terms are divisible by 3?

 

[XJellieBX]

Thank you =) I see it now.