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Regarding the question, the prime must be a natural number, so n>0. I used Mathematica to test for primes for n ranging from $1$ to $10^5$. All of them were not prime aside from $n=1$, which gives you $2$. I believe contradiction would be used for proving that there are no other primes of this form. I am just not sure where to go from there.

Thank you for your assistance.

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    Well, $3^n-1$ is even, right? – 2017-01-17
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    $3^n$ tends to be quite odd ... – 2017-01-17
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    $3^n-1$ is even for every $n$, and the only even prime is $2$. – 2017-01-17
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    LaTeX formatting would make this much more readable... – 2017-01-17
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    Also, $0$ is not a natural number... – 2017-01-17
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    I'll upvote this for exploring with Mathematica. But having seen the result you guessed, you should have written down the values of $3^2-1$, $3^3-1$ and $3^4-1$, Then you might have noticed that you were gettng even numbers - see @Reese 's answer. – 2017-01-17
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    @TheCount According to a math professor on my university, there is no official convention whether $0$ is natural or not. In general, I agree that Latex is necessary to make a question readable. But here, the original question was very easy to read. – 2017-01-17
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    @Peter: Well, LaTeX is just my opionion. As for $0$ being a natural number, it is not a convention, but a definition. We *define* natural numbers to be positive integers, hence $0$ is not a natural number. – 2017-01-17
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    @TheCount: While you are allowed to start counting with one, some of us consider zero a natural number. Zero is not a prime (zero divisors are excluded by definition) – 2017-01-17
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    **Hint** $\ x\!-\!a\mid x^n\!-\!a^n\ $ by the Factor Theorem. Put $\,x =3,\,\ a = 1\ \ $ – 2017-01-17
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    @hardmath Would you agree that $0\notin \mathbb{N}$? – 2017-01-17
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    There is no definition that $0$ is not natural either. – 2017-01-17
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    I remember the totally pointless discussion whether the year $2000$ was the beginning of the "millenium" or the year $2001$. If that would be my only problem, I would be very happy indeed. – 2017-01-17
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    @hardmath Well, as long as you are clear across the board within a document, I suppose there's no real issue, but I have always been told from a young age that $0$ is *not* natural. I just googled and was *astounded* that both are popular conventions. Good to know, I suppose. – 2017-01-17
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    An easy way to avoid confusions is to speak of non-negative integers or of positive integers. Then, everything is clear. – 2017-01-17
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    @TheCount But I do understand your point as well. It seems absurd to count, for example $0$ apples. If we count something, we always start with $1$. In some situations, it is an advantage to add the $0$, but I admit that I have no good argument for $0\in\mathbb N$, perhaps hardmath has one ? – 2017-01-17
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    @Peter I'm fine with the convention argument. Convention is a catch-all for me, even if the convention is strange or counter-intuitive. It doesn't change anything as long as one is clear. – 2017-01-17
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    More general theorem can be presented: for all *odd* $k > 2$ $ \ \ $ $k^n-1$ is not a prime :) – 2017-01-17
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    @Peter: My "argument" for including zero in the natural numbers is to supply it with an additive identity, so an algebraic conceit ([semiring](https://en.wikipedia.org/wiki/Semiring)).. – 2017-01-17
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    @Widawensen We have to rule out $n=1$ as well (or alternatively, we have to demand $k>4$) – 2017-01-17

2 Answers 2

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A good approach to problems of this form is to assume (1) that there are no primes of this sort other than the easy example, and (2) that there's a simple reason, like a single number that must divide all of them. To determine what this "simple reason" is, look at the first few examples: $2, 8, 26, 80, \ldots$. The one factor these all have in common is $2$. So it's reasonable to suppose that every example will be even; we just need to know why. But then the answer is evident: $3^n$ is a product of odd numbers, so it will always be odd. An odd number minus one is always even. And if $n > 1$, then $3^n$ will be greater than $3$, so $3^n - 1$ will always be an even number greater than two - and hence will not be prime.

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A little mistake when you say "for $n=0$ is gives you $2$", you want to say "for $n=1$.

Otherwise, notice that

$$3^n-1=0\pmod 2$$

(it is always even since $3^n$ is always odd).

So $3^n-1$ can not be a prime once it is greater than $2$. Therefore, the only case acceptable if $3^n-1=2$, which is the case for $n=1$.