Proving that $8^n - 3^n$ is divisible by $5$

Question

I really should be able to do this but I don't know why I can't figure it out.

My problem is that I have to prove $8^n - 3^n$ is divisible by $5$.

So what I did is I tried it for $n=1, n=2, n=3$ and so on and all the numbers I tried, the expression was divisible by 5.

So then what I did is I said "assume it's true for $n=g$"

So the expression turns into $8^g - 3^g$ is divisible by 5

Then I wanted to try it for $g+1$ so:

$8^{g+1} - 3^{g+1} $

$(8^g * 8) - (3^g * 3)$

Where should I go from here?

Answer 1

The induction hypothesis can be written $$ 8^g-3^g=5k $$ for some integer $k$. Therefore $8^g=3^g+5k$. Then $$ 8^{g+1}-3^{g+1}=8\cdot 8^g-3^{g+1}= 8\cdot 3^g+40k-3\cdot3^g=(8-3)\cdot 3^g+40k $$ Can you finish?

Answer 2

$8^{n+1}-3^{n+1}=8(8^{n})-3(3^{n})=8(8^{n}-3^{n})+3^{n} \times 8 - 3(3^{n})=8(8^{n}-3^{n})+3^{n} \times 5$. Could you continue from here?

Answer 3

Since $8\equiv 3\pmod 5$, the expression reduces to $3^n-3^n$ modulo 5.

If you don't have modular arithmetic available, write $8^n$ as $(3+5)^n$ and expand using the binomial theorem. The $\binom n03^n5^0$ term is just what you subtract away, and every other term has at least one factor of $5$.

Answer 4

Recall the difference of $n$th powers factorization. \begin{equation} a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n - 1}) \end{equation} Therefore \begin{equation} 8^n - 3^n = (8 - 3)(8^{n - 1} + 3 \cdot 8^{n - 2} + \cdots + 8 \cdot 3^{n - 2} + 3^{n - 1}) = 5M \end{equation} where $M = 8^{n - 1} + 3 \cdot 8^{n - 2} + \cdots + 8 \cdot 3^{n - 2} + 3^{n - 1}$. Since $M$ is an integer it is clear that $8^n - 3^n$ is divisible by 5.

Answer 5

$$8^n - 3^n = 5\sum_{k=0}^{n-1} 8^k \cdot 3^{n-k-1}$$