# Divisibility Proofs

1. Jul 8, 2012

### odolwa99

Can anyone help me confirm if I have solved this correctly?

Many thanks.

1. The problem statement, all variables and given/known data

Q. $9^n-5^n$ is divisible by 4, for $n\in\mathbb{N}_0$

3. The attempt at a solution

Step 1: For $n=1$...
$9^1-5^1=4$, which can be divided by $4$.
Therefore, $n=1$ is true...

Step 2: For $n=k$...
Assume $9^k-5^k=4\mathbb{Z}$, where $\mathbb{Z}$ is an integer...1
Show that $n=k+1$ is true...
i.e. $9^{k+1}-5^{k+1}$ can be divided by $4$
$9^{k+1}-5^{k+1}$ => $9^{k+1}-9\cdot5^k+9\cdot5^k-5^{k+1}$ => $9(9^k-5^k)+5^k(9-5)$ => $9(4\mathbb{Z})+5^k(4)$...from 1 above => $36\mathbb{Z}+5^k\cdot4$ => $4(9\mathbb{Z}+5^k)$

Thus, assuming $n=k$, we can say $n=k+1$ is true & true for $n=2,3,$... & all $n\in\mathbb{N}_0$

2. Jul 8, 2012

### micromass

That's good. Some remarks on notation:

Using $\mathbb{Z}$ is not good here. That is the symbol for the set of all integers. You should write $9^k-5^k=4z$ where z is an integer.

The => should be =

3. Jul 8, 2012

### odolwa99

Ok. Thank you.