Does there exist a positive integer $n$ such that $4^n - 1$ is not divisible by $3$? If so, what is it, and if not, prove it. This is just a recreational problem that seemed simple at first but I couldn't figure it out.
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2Use induction to show $3|(4^n-1)$ for all $n$ – Ross Millikan Apr 06 '20 at 02:47
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No such $n$ exists. $4\equiv 1 \bmod 3$, so $4^n-1\equiv 1^n-1=0 \bmod 3$, which means every number of the form $4^n-1$ is divisible by $3$.
Keith Backman
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No because
$$4^n-1=3\left(1+4+4^2+\cdots+4^{n-1}\right).$$
The way I think about this is as follows: in general we have the geometric sum $$\frac{1}{1-r}=1+r+r^2+r^3+\cdots$$
for $|r|<1$. Notice that
$$\frac{r^{n+1}}{1-r}=r^{n+1}+r^{n+2}+r^{n+3}+\cdots$$ and thus
$$\frac{1-r^{n+1}}{1-r}=1+r+r^2+\cdots+r^n.$$
This last equality is true for any real number $r\neq 1$, not just when $|r|<1$. Taking $r=4$, this gives
$$\frac{1-4^n}{-3}=1+4+4^2+\cdots+4^{n-1}$$
or rather
$$4^n-1=3(1+4+\cdots+4^{n-1}).$$
pancini
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1Or consider it as a telescoping sum $$4^n - 1 = (4^n - 4^{n-1}) + (4^{n-1} - 4^{n-2}) + \ldots + (4^1 - 4^0) = (4-1)(4^{n-1} + 4^{n-2} + \ldots + 4^0)$$ – Robert Israel Apr 06 '20 at 02:54
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If you are familiar with the binomial theorem:
$4^n-1=(3+1)^n-1 = \displaystyle \sum_{k=1}^{n} \binom{n}{k}3^{k}$.
