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I am stuck on one question:

Show that $8^n-3^n$ is divisible by $5$.

Thank you

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closed as off-topic by Jonas Meyer, N. F. Taussig, Alan, Adam Hughes, 2mkgz Mar 24 at 4:01

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What have you tried? Try factoring. –  Graphth Mar 13 '12 at 2:41
I'm agree with Graphth –  Lrrr Apr 8 '12 at 15:58
In order to receive help, it is always best form to show what you tried and not expect somebody to just give you an answer. –  agent154 Sep 20 '13 at 3:31

3 Answers 3

up vote 4 down vote accepted

If n=1, we have $8^1-3^1=5$, which is divided by $5$.

Now, assume that $8^n - 3^n$ is divisible by $5$, i.e. $8^n - 3^n = 5k$ for some $k\in N$. Then $8^n = 5k + 3^n$. Thus, for $n + 1$, we have $8^{n+1} - 3^{n+1} = 8 · 8^n - 3 · 3^n = 8(5k + 3^n) - 3 · 3^n = 8 · 5k + 5 · 3^n = 5(8k + 3^n)$. Since $8k + 3^n$ is integer, we obtain that $8^{n+1} - 3^{n+1}$ is divisible by $5$. So, we proved the induction step. Thus we proved the statement.

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Hint $\rm\ \ \ 8^{N+1}-\:3^{N+1} =\: (8-3)\:8^N + 3\:(8^N-3^N).\ $ This is prototypical telescopy.

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In case you want to improve your basic knowledge about mathematics, it is good to know about Modular arithmetic. with a little knowledge in that field you will know that $8\equiv 3 \mod{5}$.

A simpler(to understand) solution if you didn't understand module :

consider 8 = 5+3 so: $$8^n = (5+3)^n = 5^n + \dbinom{n}{1} * 5^{n-1} * 3 + ... + \dbinom{n}{n-1} * 5 * 3^{n-1} + 3^n = 5*k+3^n$$ As you can see n first term have 5 in them so : $$8^n-3^n = (5*k+3^n)-3^n = 5k$$

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@RebeccaJ.Stones This answer was posted three years ago. –  columbus8myhw Mar 24 at 2:45
Oh, I didn't realize. The question came up in the review queue, so I reviewed it. –  Rebecca J. Stones Mar 24 at 2:47
@RebeccaJ.Stones Thanks for your comment, I'll change the link :) –  Lrrr Mar 24 at 20:54

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