# Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.

I encountered a problem in a book that was designed for IMO trainees. The problem had something to do with divisibility.

Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.

Can somebody give me a hint on this problem. I know that it can be done via the principle of mathematical induction, but I am looking for some other way (that is if there is some other way)

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Hint $2^{3n}=8^n$ –  Thomas Andrews Aug 23 '12 at 16:07
Implicitly, somewhere induction will be involved. You can prove it directly using theorems, but then those theorems used induction. –  Thomas Andrews Aug 23 '12 at 16:12
For example Bidit's solution reduces the problem to $1^n-1$. It might seem silly to have to prove $1^n=1$, but formally it has to be done with induction. Many teachers would let you take $1^n=1$ for granted though (unless they are specifically looking for induction arguments...) –  rschwieb Aug 23 '12 at 16:22

Hint: Note that $8 \equiv 1~~~(\text{mod } 7)$.
So, $$2^{3n}=(2^3)^n=8^n\equiv \ldots~~~(\text{mod } 7)=\ldots~~~(\text{mod } 7)$$ Try to fill in the gaps!

Solution: Note that $8 \equiv 1~~(\text{mod } 7)$. This means that $8$ leaves a remainder of $1$ when divided by $7$. Now assuming that you are aware of some basic modular arithmetic, $$2^{3n}=(2^3)^n=8^n\equiv 1^n ~~(\text{mod } 7)=1~~(\text{mod } 7)$$ Now if $2^{3n}\equiv 1~~(\text{mod } 7)$ then it follows that,
$$2^{3n}-1=8^n-1\equiv (1-1)~~(\text{mod } 7)~\equiv 0~~(\text{mod } 7)\\ \implies 2^{3n}-1\equiv 0~~(\text{mod } 7)$$
Or in other words, $2^{3n}-1$ leaves no remainder when divided by $7$ (i.e. $2^{3n}-1$ is divisible by $7$). As desired

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So this is where I got so far from your help $$2^{3n}=(2^3)^n=8^n\equiv 1^n ~~~(\text{mod } 7)=1~~~(\text{mod } 7)$$ Which means $2^{3n} \equiv 1 ~~~(\text{mod } 7)$ ,right? Now what? –  Dan Hoise Aug 23 '12 at 16:20
So if $2^{3n}$, when divided by $7$ leaves a remainder of $1$, what remainder must $2^{3n}-1$ (which is, as you may have noticed, $1$ less than $2^{3n}$) leave when divided by 7? –  dashdart Aug 23 '12 at 16:24

HINT: $2^{3n}=8^n$, and for integers $n\ge2$ there is a well-known factorization of $x^n-y^n$.

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$2^{3n} -1 = 8^n -1 = (7+1)^n -1 =$ (By Binomial theorem)$= C^n _0.7^n + C^{n}_{1} . 7^{n-1} + \dotsb + C^n_{n-1}7 + C^n_n -1 = 7.(C^n _0.7^{n-1} + C^{n}_{1} . 7^{n-2} + \dotsb + C^n_{n-1})$

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Use Proof by induction.It's easy to show that if this holds for "n" ,then it holds for "n+1" and of course it's true for n=1

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The question asked for a way not involving induction. –  Rick Decker Sep 17 '12 at 17:55