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The question:

Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.

Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd

I realise that this is a proof by induction, and this is what I have so far:

\begin{align} f(n) & = 6^n + 8^n \\ & = (2\cdot 3)^n + (2^3)^n \\ & = 2^n \cdot 3^n + 2^{3n} \\ \end{align}

\begin{align} \text{Assume} ~ f(k) & = 6^k + 8^k \\ & = 2^k \cdot 3^k + 2^{3k} \\ f(k+1) & = 2^{k+1} \cdot 3^{k+1} + 2^{3(k+1)} \\ \end{align}

Where do I go from here?

Landuros
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  • I nominated for reopening. This question is not quite a duplicate of the linked one. In particular, there is a nice modular arithmetic argument available here which doesn't work in general. – Peter Woolfitt Dec 15 '17 at 09:34
  • Well, this is interesting. I already knew the answer to this question - I just wanted to try the Q&A answer style. I made sure that this was not a duplicate, and something people could learn from. If you feel the question was wrongly closed, please vote to reopen it, as my intentions are for the good. – Landuros Dec 15 '17 at 09:37

3 Answers3

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This is my first attempt at answering my own question. I realise the question is of a low quality, but I'm focusing on the question-and-answer aspect.

There are a few issues with the OP's attempt. Firstly, the base step was skipped. Secondly, and crucially, $f(k+1)$ is no longer odd, so it cannot be used in the inductive step.

Problem: Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.

Solution: Let $f(n)$ denote the statement \begin{align} f(n) & = 6^n + 8^n \\ \end{align}

Base Step ($n=1$): \begin{align} f(1) & = 6^1+8^1 \\ & = 6+8 \\ & = 14 \\ \end{align} $7 | 14 \implies f(1)$ holds.

Inductive Step $f(k) \to f(k+2)$: Fix some $k$, where $k$ is odd. Assume that

$$f(k) = 6^k + 8^k$$

holds true. To be proved is that

$$f(k+2) = 6^{k+2} + 8^{k+2}$$

follows. Beginning with the left-hand side of $f(k+2)$,

\begin{align} 6^{k+2} + 8^{k+2} & = 6^k \cdot6^2 + 8^k \cdot 8^2 \\ & = 6^k \cdot 36 + 8^k \cdot 64 \\ & = 6^k + 6^k \cdot 35 + 8^k + 8^k \cdot63 \\ & = 6^k + 8^k + 6^k \cdot 35 + 8^k \cdot63 \\ & = f(k) + 7(6^k\cdot5+8^k \cdot9) \end{align} by the inductive assumption, $7 | f(k)$. Clearly, $7 | (6^k\cdot5+8^k \cdot9)$, thereby showing $f(k+2)$ is true, completing the inductive step.

Conclusion: By mathematical induction, it is proved that $6^n+8^n$ is divisible by $7$ iff $n$ is odd. $\Box$

Landuros
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  • Good answer. :) – stressed out Dec 15 '17 at 09:28
  • I think this only proves one direction of the iff statement (though a simple mod 7 argument serves to quickly show the other direction (..actually mod 7 gives both directions very quickly)) – Peter Woolfitt Dec 15 '17 at 09:30
  • @PeterWoolfitt: It depends on the OP's level, I think. What if the OP doesn't know modular arithmetic yet? But you're right that he has proved only one direction and I didn't notice it was an iff statement. – stressed out Dec 15 '17 at 09:35
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$6\equiv1\pmod7, ~8\equiv-1\pmod7$, hence $$6^n+8^n\equiv1^n+(-1)^n\pmod7$$ $1^n+(-1)^n=0~\iff~n$ is odd.

QED
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Because $$6^n+8^n=(6+8)(6^{n-1}-6^{n-2}\cdot8+...-6\cdot8^{n-2}+8^{n-1}).$$

Michael Rozenberg
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